Alta - Ch. 4

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A basketball player has a 0.604 probability of making a three-point shot. What is the probability that it takes the player 5 tries to make a three-point shot? Round your answer to three decimals.

$0.015$0.015 This is an example of a geometric distribution with success probability p=0.604. Find q by q=1−p=1−0.604=0.396 Then take the equationP(X=k)=q(k−1)⋅pWhere k=5:P(X=5)=0.396(5−1)⋅0.604P(X=5)=0.396(4)⋅0.604P(X=5)≈0.015 By using a calculator or computer, the probability that it takes the player 5 tries to make a three-point shot is P(X=5)=geometpdf(0.604,5)≈0.015.

A marksman has a probability of 0.624 for hitting a certain long range target. What is the probability that it takes 4 shots for the marksman to hit the long range target? Round your answer to three decimals.

$0.033$0.033 This is an example of a geometric distribution with success probability p=0.624. Find q by q=1−p=1−0.624=0.376 Then take the equationP(X=k)=q(k−1)⋅pWhere k=4:P(X=4)=0.376(4−1)⋅0.624P(X=4)=0.376(3)⋅0.624P(X=4)≈0.033 By using a calculator or computer, the probability that it takes 4 shots for the marksman to hit the long range target is P(X=4)=geometpdf(0.624,4)≈0.033.

A weighted coin has a 0.46 probability of landing on heads. What is the probability that it takes 5 flips for the first head to occur? Round the final answer to three decimal places.

$0.039$0.039 This is an example of a geometric distribution with success probability p=0.46. Find q by q=1−p=1−0.46=0.54 Then take the equationP(X=k)=q(k−1)⋅pWhere k=5:P(X=5)=0.54(5−1)⋅0.46P(X=5)=0.54(4)⋅0.46P(X=5)=0.039 By using a calculator or computer, the probability that it takes 5 flips for the first head to occur is P(X=5)=geometpdf(0.46,5)=0.039.

A weighted coin has a 0.564 probability of landing on heads. What is the probability that it takes 4 flips for the first head to occur? Round the final answer to three decimal places.

$0.047$0.047 This is an example of a geometric distribution with success probability p=0.564. Find q by q=1−p=1−0.564=0.436 Then take the equationP(X=k)=q(k−1)⋅pWhere k=4:P(X=4)=0.436(4−1)⋅0.564P(X=4)=0.436(3)⋅0.564P(X=4)=0.047 By using a calculator or computer, the probability that it takes 4 flips for the first head to occur is P(X=4)=geometpdf(0.564,4)=0.047.

A basketball player has a 0.498 probability of making a three-point shot. What is the probability that it takes the player 4 tries to make a three-point shot? (Round your answer to three decimals if necessary.)

$0.063$0.063 This is an example of a geometric distribution with success probability p=0.498. Find q by q=1−p=1−0.498=0.502 Then take the equationP(X=k)=q(k−1)⋅pWhere k=4:P(X=4)=0.502(4−1)⋅0.498P(X=4)=0.502(3)⋅0.498P(X=4)=0.063 By using a calculator or computer, the probability that it takes the player 4 tries to make a three-point shot is P(X=4)=geometpdf(0.498,4)=0.063.

You play a casino game until you win for the first time. On average, you win for the first time on the 11th game played. What is the probability of winning the game? Round your answer to three decimal places.

$0.091$0.091 This is an example of a geometric distribution. If p represents the probability of success, then the expected number of trials is n=1p. Solving this equation for p shows that p=1n. In this case n=11, so p=111=0.091.

A restaurant gets an average of 14 calls in a 2 hour time period. What is the probability that the restaurant will get at most 2 calls in a 45 minute period? Round the final answer to three decimal places.

$0.105$0.105 The time interval of interest is 45 minutes. There is an average of 14 calls per 2 hours or 1412045=214 calls per 45 minutes. The probability can be found using the Poisson distribution with parameter λ=214. Let X be the number of calls that are received in the 45 minute time period. According to the formula, we find: P(X=x)=λxe−λx! P(X≤2)=P(X=0)+P(X=1)+P(X=2) P(X≤2)=(214)0e−2140!+(214)1e−2141!+(214)2e−2142! 0.005+0.028+0.072 ≈0.105

Nick gets an average of 18 calls during his 8 hour work day. What is the probability that Nick will get exactly 10 calls in a 5 hour portion of his work day? (Round your answer to three decimal places.)

$0.116$0.116 The time interval of interest is 5 hours. There is an average of 18 calls per 8 hours or 1885=454 calls per 5 hours. The probability can be found using the Poisson distribution with parameter λ=454. Let X be the number of calls that are received in the 5 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=10)=(454)10e−45410! ≈0.116

On average, Benjamin has noticed that 20 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probability that exactly 9 trains will pass his house in a 12 hour time period? Round your answer to three decimal places.

$0.125$0.125 The time interval of interest is 12 hours. There is an average of 20 trains per 24 hours or 202412=10 trains per 12 hours. The probability can be found using the Poisson distribution with parameter λ=10. Let X be the number of trains that pass in the 12 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=9)=109e−109! ≈0.125

Michelle gets an average of 19 calls during her 8 hour work day. What is the probability that Michelle will get exactly 6 calls in a 2 hour portion of her work day? Round your answer to three decimal places.

$0.138$0.138 The time interval of interest is 2 hours. There is an average of 19 calls per 8 hours or 1982=194 calls per 2 hours. The probability can be found using the Poisson distribution with parameter λ=194. Let X be the number of calls that are received in the 2 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=6)=(194)6e−1946! ≈0.138

Stephanie gets an average of 11 calls during her 8 hour work day. What is the probability that Stephanie will get more than 4 calls in a 2 hour portion of her work day? (Round your answer to three decimal places.)

$0.144$0.144 The time interval of interest is 2 hours. There is an average of 11 calls per 8 hours or 1182=114 calls per 2 hours. The probability can be found using the Poisson distribution with parameter λ=114. Let X be the number of calls that are received in the 2 hour time period. According to the formula, we find: P(X=x)=λxe−λx! Since it is easier to find X equal to or less than 4, we use the compliment:P(X>4)=1−(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4))P(X>4)=1−((114)0e−1140!+(114)1e−1141!+(114)2e−1142!+(114)3e−1143!+(114)4e−1144!)1−(0.064+0.176+0.242+0.222+0.152)≈0.144

On average, Andrew has noticed that 14 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probability that exactly 6 trains will pass his house in a 12 hour time period? (Round your answer to three decimal places.)

$0.149$0.149 The time interval of interest is 12 hours. There is an average of 14 trains per 24 hours or 142412=7 trains per 12 hours. The probability can be found using the Poisson distribution with parameter λ=7. Let X be the number of trains that pass in the 12 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=6)=76e−76! ≈0.149

You purchase boxes of cereal until you obtain one with the collector's toy you want. If, on average, you get the toy you want in every 6th cereal box, what is the probability of getting the toy you want in any given cereal box? Round your answer to three decimals.

$0.167$0.167 This is an example of a geometric distribution. If p represents the probability of success, then the expected number of trials is n=1p. Solving this equation for p shows that p=1n. In this case n=6, so p=16=0.167.

On average, Ruth has noticed that 19 trains pass by her house daily (24 hours) on the nearby train tracks. What is the probability that more than 4 trains will pass her house in a 4 hour time period? Round the final answer to three decimal places.

$0.214$0.214 The time interval of interest is 4 hours. There is an average of 19 trains per 24 hours or 19244=196 trains per 4 hours. The probability can be found using the Poisson distribution with parameter λ=196. Let X be the number of trains that pass in the 4 hour time period. According to the formula, we find: P(X=x)=λxe−λx! Since it is easier to find X equal to or less than 4, we use the compliment:P(X>4)=1−(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4))P(X>4)=1−((196)0e−1960!+(196)1e−1961!+(196)2e−1962!+(196)3e−1963!+(196)4e−1964!)1−(0.042+0.133+0.211+0.223+0.177)≈0.214

Teresa has a bird feeder which gets visited by an average of 14 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by exactly 1 birds in a 20 minute period during daylight hours? (Round your answer to three decimal places.)

$0.226$0.226 The time interval of interest is 20 minutes. There is an average of 14 birds per 2 hours or 1412020=73 birds per 20 minutes. The probability can be found using the Poisson distribution with parameter λ=73. Let X be the number of birds that visit the feeder in the 20 minute time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=1)=(73)1e−731! ≈0.226

On average, Mary has noticed that 18 trains pass by her house daily (24 hours) on the nearby train tracks. What is the probability that exactly 1 train will pass her house in a 3 hour time period? Round the final answer to three decimal places.

$0.237$0.237 The time interval of interest is 3 hours. There is an average of 18 trains per 24 hours or 18243=94 trains per 3 hours. The probability can be found using the Poisson distribution with parameter λ=94. Let X be the number of trains that pass in the 3 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=1)=(94)1e−941! ≈0.237

Sean gets an average of 20 calls during his 8 hour work day. What is the probability that Sean will get at most 9 calls in a 4 hour portion of his work day? Round the final answer to three decimal places.

$0.458$0.458 The time interval of interest is 4 hours. There is an average of 20 calls per 8 hours or 2084=10 calls per 4 hours. The probability can be found using the Poisson distribution with parameter λ=10. Let X be the number of calls that are received in the 4 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X≤9)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5) +P(X=6)+P(X=7)+P(X=8)+P(X=9) P(X≤9)=100e−100!+101e−101!+102e−102!+103e−103!+104e−104!+105e−105! +106e−106!+107e−107!+108e−108!+109e−109! 0.00005+0.0005+0.002+0.008+0.019+0.038+0.063+0.090+0.113+0.125 ≈0.458

Consider how the following scenario could be modeled with a binomial distribution, and answer the question that follows. 54.4% of tickets sold to a movie are sold with a popcorn coupon, and 45.6% are not. You want to calculate the probability of selling exactly 6 tickets with popcorn coupons out of 10 total tickets (or 6 successes in 10 trials). What value should you use for the parameter p?

$0.544$0.544 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a movie ticket with a popcorn coupon, so p=0.544.

Justin has a bird feeder which gets visited by an average of 18 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by more than 5 birds in a 40 minute period during daylight hours? (Round your answer to three decimal places.)

$0.554$0.554 The time interval of interest is 40 minutes. There is an average of 18 birds per 2 hours or 1812040=6 birds per 40 minutes. The probability can be found using the Poisson distribution with parameter λ=6. Let X be the number of birds that visit the feeder in the 40 minute time period. According to the formula, we find: P(X=x)=λxe−λx! Since it is easier to find X equal to or less than 5, we use the compliment:P(X>5)=1−(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5))P(X>5)=1−(60e−60!+61e−61!+62e−62!+63e−63!+64e−64!+65e−65!)1−(0.002+0.015+0.045+0.089+0.134+0.161)≈0.554

Jerry has a bird feeder which is visited by an average of 14 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by more than 2 birds in a 25 minute period during daylight hours? Be sure to round your answer to three significant figures.

$0.558$0.558 The time interval of interest is 25 minutes. There is an average of 14 birds per 2 hours or 1412025=3512 birds per 25 minutes. The probability can be found using the Poisson distribution with parameter λ=3512. Let X be the number of birds that visit the feeder in the 25 minute time period. According to the formula, we find: P(X=x)=λxe−λx! Since it is easier to find X equal to or less than 2, we use the compliment:P(X>2)=1−(P(X=0)+P(X=1)+P(X=2))P(X>2)=1−((3512)0e−35120!+(3512)1e−35121!+(3512)2e−35122!)1−(0.054+0.158+0.230)≈0.558

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.675 probability of throwing a strike for each pitch and a 0.325 probability of throwing a ball. If the softball pitcher throws 29 pitches, we want to know the probability that exactly 19 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer

$0.675$0.675 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.675.

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.692 probability of throwing a strike for each pitch and a 0.308 probability of throwing a ball. If the softball pitcher throws 29 pitches, we want to know the probability that more than 17 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

$0.692$0.692 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.692.

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.721 probability of throwing a strike for each pitch and a 0.279 probability of throwing a ball. If the softball pitcher throws 19 pitches, we want to know the probability that more than 15 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

$0.721$0.721 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.721.

A weighted die has probability 0.104 of landing on a 6. If you roll the die until you get a 6 for the first time, what is the expected number of rolls it will take? Round to the nearest integer.

$10$10 This is an example of a geometric distribution with success probability p=0.104. The expected value of the random variable of a geometric distribution is 1p=10.104=9.62. Therefore, the expected number of rolls is 10.

A casino features a game in which a weighted coin is tossed several times. The table shows the probability of each payout amount (assume that the remaining probability has a payout of 0 so that the probabilities add to 1). To the nearest dollar, what is expected payout of the game? Payout Amount $130 $2,800 $135,000 Probability 0.131 0.021 0.0003

$116\text{ dollars}$116 dollars The table shows the probability density function where the random variable takes on the values 130, 2800, and 135000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $130(0.131)+$2800(0.021)+$135000(0.0003)≈$116.

A bakery gets an average of 20 phone orders in a 2 hour time period. In order to find the probability that the bakery will get exactly 2 phone orders in a 15 minute period using the Poisson distribution, what is the time interval of interest?

$15\ \min$15 min The time interval of interest is the fixed time period for which the probability of an event is being sought. In this case, the time interval of interest is 15 minutes.

$165\text{ dollars}$165 dollars The table shows the probability density function where the random variable takes on the values 180, 1000, and 155000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $180(0.151)+$1000(0.029)+$155000(0.0007)≈$165.

A player can play a game by rolling a fair die several times. The player can win money based on the numbers rolled. The table shows the probability of each payout amount (assume that the remaining probability has a payout of 0 so that the probabilities add to 1). To the nearest dollar, what is the expected payout of the game? Payout Amount $130 $5,000 $130,000 Probability 0.171 0.026 0.0002

$178\text{ dollars}$178 dollars The table shows the probability density function where the random variable takes on the values 130, 5000, and 130000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $130(0.171)+$5000(0.026)+$130000(0.0002)≈$178.

Give the numerical value of the parameter n in the following binomial distribution scenario.The probability of buying a movie ticket with a popcorn coupon is 0.597 and without a popcorn coupon is 0.403. If you buy 18 movie tickets, we want to know the probability that no more than 13 of the tickets have popcorn coupons.Consider tickets with popcorn coupons as successes in the binomial distribution. Do not include n= in your answer.

$18$18 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, the total number of trials, or movie tickets, is n=18.

A guessing game at a casino features 50 cards labeled with the numbers 1 though 50. Four cards will be drawn without replacement and each player will guess the card numbers. The probability of each payout amount is shown in the table (assume that the remaining probability has a payout of 0 so that the probabilities add to 1). To the nearest dollar, what is the expected payout of the game? Payout Amount $100 $3,000 $150,000 Probability 0.146 0.023 0.0007

$189\text{ dollars}$189 dollars The table shows the probability density function where the random variable takes on the values 100, 3000, and 150000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $100(0.146)+$3000(0.023)+$150000(0.0007)≈$189.

$191$191 The table shows the probability density function where the random variable takes on the values 160, 3400, and 120000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $160(0.136)+$3400(0.025)+$120000(0.0007)=$191.

A restaurant gets an average of 10 phone orders in a 2 hour time period. In order to find the probability that the restaurant will get exactly 0 phone orders in a 20 minute period using the Poisson distribution, what is the time interval of interest?

$20\ \min$20 min The time interval of interest is the fixed time period for which the probability of an event is being sought. In this case, the time interval of interest is 20 minutes.

A gambling game involves a spinner with the numbers 1 through 25. To play, the player guesses which numbers the spinner will land on. The table shows the probability of each payout amount (assume that the remaining probability has a payout of 0 so that the probabilities add to 1). To the nearest dollar, what is the expected payout of the game? Payout Amount $140 $4,400 $170,000 Probability 0.116 0.021 0.0007

$228$228 The table shows the probability density function where the random variable takes on the values 140, 4400, and 170000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $140(0.116)+$4400(0.021)+$170000(0.0007)=$228.

$230$230 The table shows the probability density function where the random variable takes on the values 160, 4400, and 145000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $160(0.146)+$4400(0.024)+$145000(0.0007)=$230.

$250\text{ dollars}$250 dollars The table shows the probability density function where the random variable takes on the values 160, 4800, and 195000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $160(0.116)+$4800(0.028)+$195000(0.0005)≈$250.

$258$258 The table shows the probability density function where the random variable takes on the values 130, 4400, and 195000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $130(0.156)+$4400(0.023)+$195000(0.0007)=$258.

The probability of buying a box of cereal with a winning coupon is 0.171. If you buy boxes of this cereal until you get the winning coupon for the first time, what is the expected number of purchases it will take? Round your answer to the nearest whole number.

$6$6 This is an example of a geometric distribution with success probability p=0.171. The expected value of the random variable of a geometric distribution is 1p=10.171=5.85. Therefore, the expected number of purchases is 6.

A weighted die has probability 0.176 of landing on a 6. If you roll the die until you get a 6 for the first time, what is the expected number of rolls it will take?

$6\text{ rolls}$6 rolls This is an example of a geometric distribution with success probability p=0.176. The expected value of the random variable of a geometric distribution is 1p=10.176≈5.68. Therefore, the expected number of rolls is 6.

A guessing game at a casino features 50 cards labeled with the numbers 1 through 50. Four cards will be drawn without replacement and each player will guess the card numbers. The probability of each payout amount is shown in the table (assume that the remaining probability has a payout of 0 so that the probabilities add to 1). To the nearest dollar, what is the expected payout of the game? Payout Amount $100 $1,400 $180,000 Probability 0.106 0.023 0.0002

$79$79 The table shows the probability density function where the random variable takes on the values 100, 1400, and 180000. To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $100(0.106)+$1400(0.023)+$180000(0.0002)=$79.

The table below shows an incomplete probability density function for the discrete random variable X, the number of candy bars purchased per customer at a movie theater. What should the missing probability be? Provide the final answer as a fraction. x P(X = x) 0 1 1/4 2 1/2

$\frac{1}{4}$1/4 Remember that the probabilities in the probability density function must add up to 1. So if we let the unknown value be A, we find that A+14+12=1 So solving for A, we find that A=1/4.

Christine has a motion detector light which gets activated an average of 16 times every 2 hours during the night. In order to find the probability that the motion detector light will be activated more than 4 times in a 25 minute period during the night using the Poisson distribution, what is the average number of activations per 25 minutes? Round your answer to three decimal places.

$\lambda=3.333$λ=3.333 The ratio of the average number of activations in 2 hours is equal to the ratio of the average number of activations in 25 minutes. 2 hours is equivalent to 120 minutes. Let λ represent the average number of activations in 25 minutes, then λ25=16120. Solving this equation for λ shows that λ=16(25)120=3.333.

The table below represents the probability density function for the random variable X. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 1 1/4 3 1/4 4 1/2

$\text{std=}1.22$std=1.22 Here's the filled out table. x134P(X=x)141412x⋅P(X=x)14342(x−μ)2P(X=x)1.00.00.5 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=14+34+2=3 This value is used to compute the values in the fourth column. For example, the first entry is(1−3)2⋅14=1 Summing the values in the fourth column gives the variance:1.0+0.0+0.5=1.5 Taking the square root of the variance gives the standard deviation:σ=1.5−−−√≈1.22

A random sample of college students majoring in cinematography were surveyed about their movie habits. One question in the survey asked, "How many documentaries have you watched in the past month?" The table below represents the probability density function for the random variable X, the number of documentaries watched by cinematography majors in the past month. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 4 1/5 6 2/5 8 2/5

$\text{std=}1.50\text{ documentaries}$std=1.50 documentaries Here's the filled out table. x468P(X=x)152525x⋅P(X=x)45125165(x−μ)2P(X=x)1.1520.0641.024 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=45+125+165=6.4This value is used to compute the values in the fourth column. For example, the first entry is(4−6.4)2⋅15=1.152Summing the values in the fourth column gives the variance:1.152+0.064+1.024=2.24Taking the square root of the variance gives the standard deviation:σ=2.24−−−−√≈1.50

A random sample of high school students were surveyed about technological devices. They were asked, "How many computers do you have in your household?" The table below represents the probability density function for the random variable X, the number of computers per household. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 3 1/4 4 1/4 7 1/2

$\text{std=}1.79\text{ computers}$std=1.79 computers Here's the filled out table. x347P(X=x)141412x⋅P(X=x)34172(x−μ)2P(X=x)1.26560.39061.5313 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=34+1+72=5.25This value is used to compute the values in the fourth column. For example, the first entry is(3−5.25)2⋅14=1.2656Summing the values in the fourth column gives the variance:1.2656+0.3906+1.5313=3.1875Taking the square root of the variance gives the standard deviation:σ=3.1875−−−−−√≈1.79

Emily works at a factory that makes computer parts. The machine that produces internal hard drives results in a certain number of defects during the day. The table below represents the probability density function for the random variable X, the number of defects per day. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 4 1/6 6 1/3 9 1/2

$\text{std=}1.95\text{ defects}$std=1.95 defects Here's the filled out table. x469P(X=x)161312x⋅P(X=x)23292(x−μ)2P(X=x)1.67130.45371.6806 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=23+2+92=7.1667This value is used to compute the values in the fourth column. For example, the first entry is(4−7.1667)2⋅16=1.6713Summing the values in the fourth column gives the variance:1.6713+0.4537+1.6806=3.8056Taking the square root of the variance gives the standard deviation:σ=3.8056−−−−−√≈1.95

A random sample of college students were surveyed about technological devices. They were asked, "How many cell phone chargers do you have in your household?" The table below represents the probability density function for the random variable X, the number of cell phone chargers per household. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 4 1/4 5 1/4 9 1/2

$\text{std=}2.28\text{ chargers}$std=2.28 chargers Here's the filled out table. x459P(X=x)141412x⋅P(X=x)15492(x−μ)2P(X=x)1.89060.76562.5313 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=1+54+92=6.75This value is used to compute the values in the fourth column. For example, the first entry is(4−6.75)2⋅14=1.8906Summing the values in the fourth column gives the variance:1.8906+0.7656+2.5313=5.1875Taking the square root of the variance gives the standard deviation:σ=5.1875−−−−−√≈2.28

A random sample of finance majors taking an accounting course were surveyed after completing the final exam. They were asked, "How many times did you review your final exam before handing it in to the professor?" The table below represents the probability density function for the random variable X, the number of times students reviewed their exam before handing it in. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/5 1 1/5 5 1/5 6 2/5

$\text{std=}2.58\text{ times}$std=2.58 times Here's the filled out table. x0156P(X=x)15151525x⋅P(X=x)0151125(x−μ)2P(X=x)2.5921.3520.3922.304 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=15+1+125=3.6This value is used to compute the values in the fourth column. For example, the first entry is(0−3.6)2⋅15=2.592Summing the values in the fourth column gives the variance:2.592+1.352+0.392+2.304=6.64Taking the square root of the variance gives the standard deviation:σ=6.64−−−−√≈2.58

Barbara is a quality control inspector at a shoe factory. At the end of each day, she checks the number of imperfections found in sneakers. The table below represents the probability density function for the random variable X, the number of imperfections found in sneakers per day. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/4 2 1/4 4 1/4 7 1/4

$\text{std=}2.59\text{ imperfections}$std=2.59 imperfections Here's the filled out table. x0247P(X=x)14141414x⋅P(X=x)012174(x−μ)2P(X=x)2.64060.39060.14063.5156 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=12+1+74=3.25This value is used to compute the values in the fourth column. For example, the first entry is(0−3.25)2⋅14=2.6406Summing the values in the fourth column gives the variance:2.6406+0.3906+0.1406+3.5156=6.6874Taking the square root of the variance gives the standard deviation:σ=6.6874−−−−−√≈2.59

Andrew is a quality control inspector at a clothing factory. At the end of each day, he checks the number of imperfections found in cotton sweaters. The table below represents the probability density function for the random variable X, the number of imperfections found in cotton sweaters per day. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/6 1 1/6 4 1/3 7 1/3

$\text{std=}2.67\text{ imperfections}$std=2.67 imperfections Here's the filled out table. x0147P(X=x)16161313x⋅P(X=x)0164373(x−μ)2P(X=x)2.44911.3380.00933.3426 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=16+43+73=3.8333This value is used to compute the values in the fourth column. For example, the first entry is(0−3.8333)2⋅16=2.4491Summing the values in the fourth column gives the variance:2.4491+1.338+0.0093+3.3426=7.139Taking the square root of the variance gives the standard deviation:σ=7.139−−−−√≈2.67

A survey asked a random sample of statistics majors taking a probability course how many questions were marked incorrect on their last chapter test. The table below represents the probability density function for the random variable X, the number of questions marked incorrect. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 3 1/5 4 1/5 9 3/5

$\text{std=}2.71\text{ questions}$std=2.71 questions Here's the filled out table. x349P(X=x)151535x⋅P(X=x)3545275(x−μ)2P(X=x)2.8881.5682.904 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=35+45+275=6.8This value is used to compute the values in the fourth column. For example, the first entry is(3−6.8)2⋅15=2.888Summing the values in the fourth column gives the variance:2.888+1.568+2.904=7.36Taking the square root of the variance gives the standard deviation:σ=7.36−−−−√≈2.71

During a bowling league tournament, the number of times that teams scored a strike every ten minutes was recorded by a scorekeeper. The table below represents the probability density function for the random variable X, the number of strikes every ten minutes. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/6 3 1/6 4 1/3 8 1/3

$\text{std=}2.81\text{ strikes}$std=2.81 strikes Here's the filled out table. x0348P(X=x)16161313x⋅P(X=x)0124383(x−μ)2P(X=x)3.3750.3750.08334.0833 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=12+43+83=4.5This value is used to compute the values in the fourth column. For example, the first entry is(0−4.5)2⋅16=3.375Summing the values in the fourth column gives the variance:3.375+0.375+0.0833+4.0833=7.9166Taking the square root of the variance gives the standard deviation:σ=7.9166−−−−−√≈2.81

A random sample of college students majoring in literature were surveyed about their reading habits. One question in the survey asked, "How many novels have you read in the past month?" The table below represents the probability density function for the random variable X, the number of novels read by literature majors in the past month. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 1 1/5 6 2/5 9 2/5

$\text{std=}2.93\text{ novels}$std=2.93 novels Here's the filled out table. x169P(X=x)152525x⋅P(X=x)15125185(x−μ)2P(X=x)5.4080.0163.136 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=15+125+185=6.2This value is used to compute the values in the fourth column. For example, the first entry is(1−6.2)2⋅15=5.408Summing the values in the fourth column gives the variance:5.408+0.016+3.136=8.56Taking the square root of the variance gives the standard deviation:σ=8.56−−−−√≈2.93

The table below represents the probability density function for the random variable X. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/6 4 1/3 8 1/2

$\text{std=}2.98$std=2.98 Here's the filled out table. x048P(X=x)161312x⋅P(X=x)0434(x−μ)2P(X=x)4.74070.59263.5556 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=43+4=5.3333This value is used to compute the values in the fourth column. For example, the first entry is(0−5.3333)2⋅16=4.7407Summing the values in the fourth column gives the variance:4.7407+0.5926+3.5556=8.8889Taking the square root of the variance gives the standard deviation:σ=8.8889−−−−−√≈2.98

The table below represents the probability density function for the random variable X. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/5 6 2/5 9 2/5

$\text{std=}3.29$std=3.29 Here's the filled out table. x069P(X=x)152525x⋅P(X=x)0125185(x−μ)2P(X=x)7.20.03.6 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=125+185=6This value is used to compute the values in the fourth column. For example, the first entry is(0−6)2⋅15=7.2Summing the values in the fourth column gives the variance:7.2+0.0+3.6=10.8Taking the square root of the variance gives the standard deviation:σ=10.8−−−−√≈3.29

During a college football national championship, the number of times football teams scored a touchdown was recorded by a scorekeeper every quarter. The table below represents the probability density function for the random variable X, the number of touchdowns per quarter. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 1 1/5 2 1/5 8 1/5 9 2/5

$\text{std=}3.54\text{ touchdowns}$std=3.54 touchdowns Here's the filled out table. x1289P(X=x)15151525x⋅P(X=x)152585185(x−μ)2P(X=x)4.6082.8880.9684.096 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=15+25+85+185=5.8This value is used to compute the values in the fourth column. For example, the first entry is(1−5.8)2⋅15=4.608Summing the values in the fourth column gives the variance:4.608+2.888+0.968+4.096=12.56Taking the square root of the variance gives the standard deviation:σ=12.56−−−−√≈3.54

The table below represents the probability density function for the random variable X. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/6 1 1/3 8 1/2

$\text{std=}3.68$std=3.68 Here's the filled out table. x018P(X=x)161312x⋅P(X=x)0134(x−μ)2P(X=x)3.12963.70376.7222 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=13+4=4.3333This value is used to compute the values in the fourth column. For example, the first entry is(0−4.3333)2⋅16=3.1296Summing the values in the fourth column gives the variance:3.1296+3.7037+6.7222=13.5555Taking the square root of the variance gives the standard deviation:σ=13.5555−−−−−−√≈3.68

Identify the parameters p and n in the following binomial distribution scenario. A basketball player has a 0.404 probability of making a free throw and a 0.596 probability of missing. If the player shoots 21 free throws, we want to know the probability that he makes no more than 6 of them. (Consider made free throws as successes in the binomial distribution.)

$p=0.404,\ n=21$p=0.404, n=21 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a made free throw, so p=0.404. The total number of trials, or free throws, is n=21.

Consider how the following scenario could be modeled with a binomial distribution, and answer the question that follows. The probability of buying a movie ticket with a popcorn coupon is 0.546 and without a popcorn coupon is 0.454. If you buy 27 movie tickets, we want to know the probability that exactly 15 of the tickets have popcorn coupons. (Consider tickets with popcorn coupons as successes in the binomial distribution.) What value should you use for the parameter p?

$p=0.546$p=0.546 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a movie ticket with a popcorn coupon, so p=0.546.

Identify the parameters p and n in the following binomial distribution scenario. The probability of winning an arcade game is 0.718 and the probability of losing is 0.282. If you play the arcade game 20 times, we want to know the probability of winning more than 15 times. (Consider winning as a success in the binomial distribution.)

$p=0.718,\ n=20$p=0.718, n=20 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is winning a game, so p=0.718. The total number of trials, or games, is n=20.

On average, Roger has noticed that 9 runners pass by his house each day. What is the probability that exactly 5 runners will pass his house in a day? (Answers are rounded to the thousandths digit.)

0.061 The time interval of interest is a day. The average number is 9 per day, so if X is the number of runners that Roger sees in a day, then X∼Po(λ) where λ=9. We want to know the probability of 5 runners in a day, which means x=5. According to the formula, we find P(X=x)=λxe−λx!=95e−95!≈0.061

According to a 2011 Pew Research poll, cell owners make or receive an average of 12 phone calls each day. Let X = the number of phone calls that smartphone users send or receive per day. The random variable X has a Poisson distribution: X ~ P(12). What is the probability that a smartphone user makes or receives exactly 20 phone calls in 2 days? (Round to the thousandths place.)

0.062 To find the probability of a Poisson Distribution, use the formula: f(x)=e−μ⋅μxx! ...where μ is the average, and x is the number of successes the question is asking about. In this question, the value of x is 20 because that is the number of phone calls (or successes) that the question is asking about. However, the interval is changed from 1 day to 2 days, so μ should be doubled. The average number of phone calls per day is 12, and 12⋅2=24. So the new value of μ is 24. Plugging these values into the equation, you get: f(x)=e−24⋅242020! ...which simplifies to 0.062, rounded to the thousandths digit.

According to a recent poll, smart phone users receive an average of 4 phone calls each day. Let X = the number of phone calls that smartphone users receive per day. The random variable X has a Poisson distribution: X ~ P(4). What is the probability that a smartphone users receives exactly 3 phone calls per day? (Rounded to the thousandths place.)

0.195 To find the probability of a Poisson Distribution, use the formula: f(x)=e−μ⋅μxx! ...where μ is the average, and x is the number of successes the question is asking about. In this question, μ is 4, the average number of phone calls smartphone users receive in a day. The value of x is 3 because that is the number of phone calls (or successes) that the question is asking about. Plugging these values into the equation, you get: f(x)=e−4⋅433! ...which simplifies to 0.195, rounded to the thousandths digit.

65% of the people in Missouri pass the driver's test on the first attempt. A group of 7 people took the test. What is the probability that at most 3 in the group pass their driver's tests in their first attempt? Round your answer to three decimal places. Remember: 65% = 0.65.

0.200 The parameters of this binomial experiment are: n = 7 trials p = 0.65 x = at most 3 successes This means that x takes on more than one value. At most 3 means that x can be 0,1,2, or 3 successes. Therefore, the binomial probability is: P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) =7C0(0.65)0(1−0.65)7−0+7C1(0.65)1(1−0.65)7−1+7C2(0.65)2(1−0.65)7−2+7C3(0.65)3(1−0.65)7−3 =7!0!(7−0)!(0.65)0(1−0.65)7+7!1!(7−1)!(0.65)1(1−0.65)6+7!2!(7−2)!(0.65)2(1−0.65)5+7!3!(7−3)!(0.65)3(1−0.65)4 ≈0.001+0.008+0.047+0.144 =0.200 So, the probability that at least 3 people pass their driver's test on their first attempt is approximately 0.200 or 20.0%.

Jamie is practicing free throws before her next basketball game. The probability that she makes each shot is 0.6. If she takes 10 shots, what is the probability that she makes exactly 7 of them? Round your answer to three decimal places.

0.215 Here, the parameters are x=7, p=0.6, and n=10. Plugging this in to the formula for B(x;n,p), we find B(x;n,p)=nCx⋅px⋅(1−p)n−x=10C7⋅(0.6)7⋅(1−0.6)3=10!3!7!⋅(0.6)7⋅(0.4)3≈120⋅(0.028)⋅(0.064)≈0.215 So the probability of making exactly 7 shots is approximately 0.215.

According to a recent poll, smart phone users receive an average of 4 phone calls each day. Let X = the number of phone calls that smartphone users receive per day. The random variable X has a Poisson distribution: X ~ P(4). What is the probability that a smartphone user receives at most 2 phone calls per day? (Round to the thousandths place.)

0.238 To find the probability of a Poisson Distribution, use the formula: f(x)=e−μ⋅μxx! ...where μ is the average, and x is the number of successes the question is asking about. In this question, μ is 4, the average number of phone calls smartphone users receive in a day. The value of x is 2 because that is the number of phone calls (or successes) that the question is asking about. But since the question asks for "at most 2," you must find the probility of P(X=0)+P(X=1)+P(X=2), all of the possible values of 2 or less. Plugging these values into the equation, you get: f(x)f(x)=P(X=0)+P(X=1)+P(X=2)=e−4⋅400!+e−4⋅411!+e−4⋅422!≈0.018316+0.073263+0.146525 ...which simplifies to 0.238, rounded to the thousandths digit.

A softball pitcher has a 0.42 probability of throwing a strike for each pitch. If the softball pitcher throws 20 pitches, what is the probability that exactly 7 of them are strikes? Round your answer to three decimal places.

1$0.150$0.150 In this case, we have that x=7, n=20, and p=0.42. Thus, the probability can be found by, P(X=7)=20C7⋅0.427⋅0.5813=20!7!13!⋅0.427⋅0.5813≈77520⋅0.0023⋅0.0008≈0.15

The probability of buying a movie ticket with a popcorn coupon is 0.717. If you buy 21 movie tickets, what is the probability that exactly 14 of the tickets have popcorn coupons? (Round your answer to 3 decimal places if necessary.)

1$0.160$0.160 In this case, we have that x=14, n=21, and p=0.717. Thus, the probability can be found by, P(X=14)=21C14⋅0.71714⋅0.2837=21!14!7!⋅0.71714⋅0.2837≈0.160

65% of the people in Missouri pass the driver's test on the first attempt. A group of 5 people took the test. What is the probability that less than 3 in the group pass their driver's tests in their first attempt? Round your answer to three decimal places. Remember: 65% = 0.65.

1$0.235$0.235 The parameters of this binomial experiment are: n=5 trials p=0.65 x= less than 3 successes This means that x takes on more than one value. Less than 3 means that x can be 0,1, or 2 successes. Now we have that P(X=0)=5C0⋅0.650⋅0.355=5!0!5!⋅0.650⋅0.355≈1⋅1⋅0.0053≈0.005 P(X=1)=5C1⋅0.651⋅0.354=5!1!4!⋅0.651⋅0.354≈5⋅0.65⋅0.015≈0.049 and P(X=2)=5C2⋅0.652⋅0.353=5!2!3!⋅0.652⋅0.353≈10⋅0.4225⋅0.0429≈0.181 So P(X<3)=0.005+0.049+0.181=0.235

A basketball player has a 0.603 probability of making a free throw. If the player shoots 10 free throws, what is the probability that she makes exactly 6 of them? (Round your answer to 2 decimal places if necessary.)

1$0.25$0.25 In this case, we have that x=6, n=10, and p=0.603. Thus, the probability can be found by,B(6;10,0.603)=10C6⋅0.6036⋅0.3974=10!6!4!⋅0.6036⋅0.3974≈210⋅0.0481⋅0.0248≈0.25

65% of the people in Missouri pass the driver's test on the first attempt. A group of 7 people took the test. Which of the following equations correctly calculate the probability that at least 3 in the group pass their driver's tests in their first attempt? Insert the correct symbol to represent the probability.

1$<$< This probability could be represented by P(X≥3) or 1−P(X<3).

A softball pitcher has a 0.507 probability of throwing a strike for each pitch. If the softball pitcher throws 15 pitches, what is the probability that exactly 8 of them are strikes? Insert the correct symbol to represent this probability.

1$=$= This probability is represented by the notation P(X=8).

A softball pitcher has a 0.487 probability of throwing a strike for each pitch. If the softball pitcher throws 29 pitches, what is the probability that no more than 14 of them are strikes? Insert the correct symbol to represent this probability.

1$\le$≤ This probability could be represented by P(X≤14) or P(X<15)

Using the same scenario, what is the standard deviation for the number of matches that they win in the tournament? The Stomping Elephants volleyball team plays 30 matches in a week-long tournament. On average, they win 4 out of every 6 matches.

6.67−−−−√ The standard deviation of a binomial distribution is the square root of the variance. The variance is the product of n, the total number of repeated trials or events, p, the probability of a success, and q, which is (1−p). variance = σ2=np(1−p) standard deviation = variance−−−−−−−√ In this scenario, the number of trials is 30 (the total number of matches played), which is n. The probability of a success (winning a match) is 46. So, the value for (1−p) is 26. The variance is the product of these values. variance = σ2=(30)(46)(26) So the variance is approximately 6.67, and the standard deviation is the square root of that, 6.67−−−−√.

Correct answer: P(X≥3)=P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7) P(X≥3)=1−[P(X=2)+P(X=1)+P(X=0)] We need to find the probability of 3,4,5,6, and 7 friends passing their driver's tests in their first attempt. There are two different ways to find this probability: Sum the probabilities of each case. Find the probability of 0,1, and 2 friends passing their driver's tests in their first attempt. Add the probabilities and subtract the sum from 1 to use the cumulative rule.

The table below shows an incomplete probability density function for the number of errors, X, found in a machine during a monthly quality control inspection at a textile factory. What should the missing probability be? Provide the final answer as a fraction. x P(X = x) 0 1/2 1 1/3 2

Correct answers:$\frac{1}{6}$16 Remember that the probabilities in the probability density function must add up to 1. So if we let the unknown value be A, we find that 12+13+A=1 So solving for A, we find that A=16.

Suppose a dice is rolled 5 times. A "success" is determined as rolling a 4. What is the probability of getting at most two successes?

The parameters of this binomial experiment are: n = 5 trials p = 16≈0.167 x = at most 2 successes This means that x takes on more than one value. At most 2 means that x can be 0,1, or 2 successes. So, we need to find P(X=0),P(X=1) and P(X=2) , and sum them. Therefore, the probability is: P(X≤2)=P(X=2)+P(X=1)+P(X=0)≈5C2(0.167)2(1−0.167)5−2+5C1(0.167)1(1−0.167)5−1+5C0(0.167)0(1−0.167)5−0≈5!2!(5−2)!⋅(0.167)2(0.833)3+5!1!(5−1)!⋅(0.167)(0.833)4+5!0!(5−0)!⋅1⋅(0.833)5≈5!2!3!(0.167)2(0.833)3+5!1!4!(0.167)(0.833)4+5!0!5!⋅(0.833)5≈10(0.016)+5(0.080)+1(0.401)≈0.961 So, the probability of rolling 4 at most two times in 5 trials is approximately 0.961 or 96.1%.

You purchase boxes of cereal until you obtain one with the collector's toy you want. If, on average, you get the toy you want in every 11th cereal box, what is the probability of getting the toy you want in any given cereal box? (Round your answer to three decimals if necessary.)

The probability of winning on a slot machine game is 0.215. If you play the slot machine until you win for the first time, what is the expected number of games it will take?

$5$5 This is an example of a geometric distribution with success probability p=0.215. The expected value of the random variable of a geometric distribution is 1p=10.215=4.65. Therefore, the expected number of games is 5.

A restaurant gets an average of 12 phone orders in a 2 hour time period. In order to find the probability that the restaurant will get exactly 2 phone orders in a 30 minute period using the Poisson distribution, what is the average number of phone orders per 30 minutes?

$\lambda=3$λ=3 The ratio of the average number of phone orders in 2 hours is equal to the ratio of the average number of phone orders in 30 minutes. Let λ represent the average number of phone orders in 30 minutes, then λ30=12120. Solving this equation for λ shows that λ=12(30)120=3.

Elizabeth gets an average of 10 emails during her 8 hour work day. In order to find the probability that Elizabeth will get more than 3 emails in a 3 hour portion of her work day using the Poisson distribution, what is the average number of emails received per 3 hours? Round your answer to two decimal places.

$\lambda=3.75$λ=3.75 The ratio of the average number of emails in 8 hours is equal to the ratio of the average number of emails in 3 hours. Let λ represent the average number of emails in 3 hours, then λ3=108. Solving this equation for λ shows that λ=10(3)8=3.75.

On average, Ruth has noticed that 20 trucks pass by her apartment daily (24 hours). In order to find the probability that more than 4 trucks will pass her apartment in a 6 hour time period using the Poisson distribution, what is the average number of trucks per 6 hours?

$\lambda=5$λ=5 The ratio of the average number of trucks in 24 hours is equal to the ratio of the average number of trucks in 6 hours. Let λ represent the average number of trucks in 6 hours, then λ6=2024. Solving this equation for λ shows that λ=20(6)24=5.

The Stomping Elephants volleyball team plays 30 matches in a week-long tournament. On average, they win 4 out of every 6 matches. What is the mean for the number of matches that they win in the tournament?

20 The mean of a binomial distribution is the product of p, the probability of a success, and n, the total number of repeated trials or events. mean = μ=np In this scenario, the number of trials is 30 (the total number of matches played), which is n. The probability of a success (winning a match) is 46. So, the mean of the binomial distribution B(30,46) is: μ=(30)(46)=20

Which are properties of a random variable? Select all that apply.

Random variables are denoted with an upper case letter. Random variables are not represented by numerical values. Notation for a random variable is an upper case letter. These variables are usually represented with words, and do not have numerical values. Recall that discrete data are data that you can count, whether it is random variable or not.

Standard deviation: a number that measures how far data values or probability distributions are from their mean, denoted by a Greek letter (σ) Discrete random variable: variables whose possible values are a list of distinct values Probability distribution: contains the probability of each possible outcome for a discrete random variable and always adds up to 1.0

Identify the parameters p and n in the following binomial distribution scenario. Jack, a bowler, has a 0.38 probability of throwing a strike and a 0.62 probability of not throwing a strike. If Jack bowls 20 times, he wants to know the probability that he throws more than 10 strikes. (Consider a strike a success in the binomial distribution.)

p=0.38,n=20 In a binomial distribution, there are only two possible outcomes. p denotes the probability of the event or trial resulting in a success. In this scenario, it would be the probability of Jack bowling a strike, which is 0.38. The total number of repeated and identical events or trials is denoted by n. In this scenario, Jack bowls a total of 20 times, so n=20.

Raymond has a motion detector light which gets activated an average of 15 times every 2 hours during the night. In order to find the probability that the motion detector light will be activated exactly 8 times in a 55 minute period during the night using the Poisson distribution, what does the random variable X represent?

the number of activations in 55 minutes The random variable X represents the number of occurrences of an event in the time interval of interest. The time interval of interest is the fixed time period for which the probability of an event is being sought. In this case, the time interval of interest is 55 minutes. The random variable X is the number of activations that occur in the 55 minute time period.

Which of the following tables shows a valid probability density function? Select all correct answers.

xP(X=x)018114258 xP(X=x)00.1310.0920.4530.2740.06 xP(X=x)012118238 Remember that a valid probability density function has all of its probabilities between 0 and 1, inclusive, and the sum of the probabilities equal 1.

For the below problem, which values would you fill in the blanks of the function B(x;n,p)? The probability of saving a penalty kick from the opposing team is 0.617 for a soccer goalie. If 7 penalty kicks are shot at the goal, what is the probability that the goalie will save 5 of them?

B(5;7,0.617) The parameters of a binomial distribution are: n = the number of trials x = the number of successes in the whole experiment p = the probability of a success The parameters should be in the order of x, n, p in the binomial function B(x;n,p). So, in this case, you should input B(5;7,0.617).

The table below shows a probability density function for a discrete random variable X. What is the probability that X is 2? x P(X = x) 0 1/12 1 5/12 2 1/12 3 1/6 4 1/6 5 1/12

1/12 Find the row where x = 2 (the third row in the table). The probability that X is 2 can be found in the righthand column of the table, in this row. So, the probability that X is 2 is 1/12.

The National Coffee Association claimed that in 2010, Americans drank an average of 3 cups of coffee per day. Let X = the number of cups of coffee Joe drinks in a day. The random variable X has a Poisson distribution: X∼Po(3) . What is the probability that he drinks 2 or fewer cups of coffee in a day? (Answers are rounded to the thousandths digit.)

0.423 The average is 3, so the parameter is λ=3. Remember that the formula is P(X=x)=λxe−λx! The probability of drinking 2 or fewer cups isP(X≤2)=P(X=0)+P(X=1)+P(X=2)=30e−30!+31e−31!+32e−32!≈0.050+0.149+0.224≈0.423

A softball pitcher has a 0.431 probability of throwing a strike for each pitch. If the softball pitcher throws 22 pitches, what is the probability that exactly 12 of them are strikes? Round your answer to 2 decimal places.

1$0.09$0.09 In this case, we have that x=12, n=22, and p=0.431. Thus, the probability can be found by,B(12;22,0.431)=22C12⋅0.43112⋅0.56910=22!12!10!⋅0.43112⋅0.56910≈646646⋅0.00004⋅0.00356≈0.09

An average person can type 40 words per minute. In order to find the probability that a person will type more than 100 words in 3 minutes using the Poisson distribution, what is the average number of words per 3 minutes?

120 words Since the mean, μ, is equal to np, you need to determine the values of n (the average over the whole interval) and p (the ratio of the chosen interval to the entire interval). The average number of words per minute is given as 40. So, n=40. The entire time interval the question asks for is 3 minutes. This interval is 3 times as long as the original. So, p=3. Substitute the values for n and p in the equation to find the mean. μ=np=(40)(3)=120 So, the mean is 120 words per 3 minutes.

The table below represents the probability density function for the random variable X. Find the standard deviation of X. x P(X = x) 0 1/5 1 2/5 8 2/5

3.61 Here's the filled out table. x018P(X=x)152525x⋅P(X=x)025165(x−μ)2P(X=x)2.5922.7047.744 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=25+165=3.6This value is used to compute the values in the fourth column. For example, the first entry is(0−3.6)2⋅15=2.592Summing the values in the fourth column and taking the square root gives the standard deviation:σ=2.592+2.704+7.744−−−−−−−−−−−−−−−−−√=13.04−−−−√≈3.61

Poisson Distribution: a discrete probability distribution that gives the probability of a number of events occurring in a fixed interval of time or space

Suppose a dice is rolled 5 times. A "success" is determined as rolling a 4. What is the probability of getting at least two successes?

The parameters of this binomial experiment are: n = 5 trials p = 16≈0.167 x = at least 2 successes This means that x takes on more than one value. At least 2 means that x can be 2,3,4, or 5 successes. So, we can either sum each of those probabililties or use the complement rule to make the calculations a little easier. Therefore, the probability is: P(X≥2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)=1−[P(X=1)+P(X=0)]≈1−[5C1(0.167)1(1−0.167)5−1+5C0(0.167)0(1−0.167)5−0]≈1−[5!1!(5−1)!⋅(0.167)(0.833)4+5!0!(5−0)!⋅1⋅(0.833)5]≈1−[5!1!4!(0.167)(0.833)4+5!0!5!⋅(0.833)5]≈1−[5(0.08041)+1(0.4011)]≈1−0.8031≈0.1969 So, the probability of rolling 4 at least two times in 5 trials is approximately 0.1969 or 19.69%.

Suppose a dice is rolled 5 times. A "success" is determined as rolling a 4. What is the probability of getting exactly two successes?

This is a binomial experiment because there are only two outcomes - a "success" (rolling a 4), or a "failure" (not rolling a 4). n = 5 trials x = 2 successes p = 16≈0.167 (The probability of rolling a number on a fair die is 16, because there are 6 faces.) Therefore, substituting the values for n, x, and p, the probability is: B(2;5,0.167)=5C2⋅(0.167)2⋅(1−0.167)5−2=5!2!(5−2)!⋅(0.167)2⋅(0.833)5−2=5!2!3!⋅(0.167)2⋅(0.833)3≈10⋅(0.028)⋅(0.578)≈0.161 So, the probability of rolling 4 exactly two times in 5 trials is 0.161 or 16.1%.

Rae packs an average of 45 boxes during her 3 hour shift. In order to find the probability that Rae will pack at least 55 boxes in a 4 hour shift using the Poisson distribution, what does the random variable X represent?

the number of boxes she packs in 4 hours The random variable X represents the number of occurrences of an event in the time interval of interest. The time interval of interest is the fixed time period for which the probability of an event is being sought. In this case, the time interval of interest is 4 hours. The random variable X is the number of boxes that Rae packs in that amount of time.

A basketball player has a 0.542 probability of making a three-point shot. What is the probability that it takes the player 5 tries to make a three-point shot? Round your answer to three decimals.

$0.024$0.024 This is an example of a geometric distribution with success probability p=0.542. Find q by q=1−p=1−0.542=0.458 Then take the equationP(X=k)=q(k−1)⋅pWhere k=5:P(X=5)=0.458(5−1)⋅0.542P(X=5)=0.458(4)⋅0.542P(X=5)≈0.024 By using a calculator or computer, the probability that it takes the player 5 tries to make a three-point shot is P(X=5)=geometpdf(0.542,5)≈0.024.

A department store gets an average of 16 calls in a 2 hour time period. What is the probability that the department store will get at most 3 calls in a 55 minute period? Round the final answer to three decimal places.

$0.066$0.066 The time interval of interest is 55 minutes. There is an average of 16 calls per 2 hours or 1612055=223 calls per 55 minutes. The probability can be found using the Poisson distribution with parameter λ=223. Let X be the number of calls that are received in the 55 minute time period. According to the formula, we find: P(X=x)=λxe−λx! P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) P(X≤3)=(223)0e−2230!+(223)1e−2231!+(223)2e−2232!+(223)3e−2233! 0.0006+0.004+0.018+0.043 ≈0.066

A weighted coin has probability 0.342 of landing heads. What is the probability that it takes 3 flips for the first head to occur? Round your answer to three decimals.

$0.148$0.148 This is an example of a geometric distribution with success probability p=0.342. Find q by q=1−p=1−0.342=0.658 Then take the equationP(X=k)=q(k−1)⋅pWhere k=3:P(X=3)=0.658(3−1)⋅0.342P(X=3)=0.658(2)⋅0.342P(X=3)≈0.148 By using a calculator or computer, the probability that it takes 3 flips for the first head to occur is P(X=3)=geometpdf(0.342,3)≈0.148.

Joseph gets an average of 13 calls during his 8 hour work day. What is the probability that Joseph will get exactly 4 calls in a 3 hour portion of his work day? Round the final answer to three decimal places.

$0.180$0.180 The time interval of interest is 3 hours. There is an average of 13 calls per 8 hours or 1383=398 calls per 3 hours. The probability can be found using the Poisson distribution with parameter λ=398. Let X be the number of calls that are received in the 3 hour time period. According to the formula, we find: P(X=x)=λxe−λx! P(X=4)=(398)4e−3984! ≈0.180

A basketball player has a 0.462 probability of making a three-point shot. What is the probability that it takes the player 2 tries to make a three-point shot? (Round your answer to three decimals if necessary.)

$0.249$0.249 This is an example of a geometric distribution with success probability p=0.462. Find q by q=1−p=1−0.462=0.538 Then take the equationP(X=k)=q(k−1)⋅pWhere k=2:P(X=2)=0.538(2−1)⋅0.462P(X=2)=0.538(1)⋅0.462P(X=2)=0.249 By using a calculator or computer, the probability that it takes the player 2 tries to make a three-point shot is P(X=2)=geometpdf(0.462,2)=0.249.

On average, Charles has noticed that 13 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probability that more than 5 trains will pass his house in a 8 hour time period? (Round your answer to three decimal places.)

$0.269$0.269 The time interval of interest is 8 hours. There is an average of 13 trains per 24 hours or 13248=133 trains per 8 hours. The probability can be found using the Poisson distribution with parameter λ=133. Let X be the number of trains that pass in the 8 hour time period. According to the formula, we find: P(X=x)=λxe−λx! Since it is easier to find X equal to or less than 5, we use the compliment:P(X>5)=1−(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5))P(X>5)=1−((133)0e−1330!+(133)1e−1331!+(133)2e−1332!+(133)3e−1333!+(133)4e−1334!+(133)5e−1335!)1−(0.013+0.057+0.123+0.178+0.193+0.167)≈0.269

Nick has a bird feeder which gets visited by an average of 17 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by at most 5 birds in a 25 minute period during daylight hours? Round the final answer to three decimal places.

$0.852$0.852 The time interval of interest is 25 minutes. There is an average of 17 birds per 2 hours or 1712025=8524 birds per 25 minutes. The probability can be found using the Poisson distribution with parameter λ=8524. Let X be the number of birds that visit the feeder in the 25 minute time period. According to the formula, we find: P(X=x)=λxe−λx! P(X≤5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5) P(X≤5)=(8524)0e−85240!+(8524)1e−85241!+(8524)2e−85242!+(8524)3e−85243!+(8524)4e−85244!+(8524)5e−85245! 0.029+0.103+0.181+0.214+0.190+0.135 ≈0.852

The probability of winning on a slot machine game is 0.035. If you play the slot machine until you win for the first time, what is the expected number of games it will take?

$29\ games$29 games This is an example of a geometric distribution with success probability p=0.035. The expected value of the random variable of a geometric distribution is 1p=10.035≈28.57. Therefore, the expected number of games is 29.

$\lambda=2.5$λ=2.5 The ratio of the average number of phone orders in 2 hours is equal to the ratio of the average number of phone orders in 15 minutes. Let λ represent the average number of phone orders in 15 minutes, then λ15=20120. Solving this equation for λ shows that λ=20(15)120=2.5

A florist shop gets an average of 18 phone orders in a 2 hour time period. In order to find the probability that the florist shop will get at most 4 phone orders in a 30 minute period using the Poisson distribution, what is the average number of phone orders per 30 minutes? Round your answer to one decimal place.

$\lambda=4.5$λ=4.5 The ratio of the average number of phone orders in 2 hours is equal to the ratio of the average number of phone orders in 30 minutes. Let λ represent the average number of phone orders in 30 minutes, then λ30=18/120. Solving this equation for λ shows that λ=18(30)/120=4.5.

The table below represents the probability density function for the random variable X. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 0 1/5 2 2/5 6 2/5

$\text{std=}2.40$std=2.40 Here's the filled out table. x026P(X=x)152525x⋅P(X=x)045125(x−μ)2P(X=x)2.0480.5763.136 Note that the mean (also called expected value) μ is the sum of the entries in the third column:μ=45+125=3.2This value is used to compute the values in the fourth column. For example, the first entry is(0−3.2)2⋅15=2.048Summing the values in the fourth column gives the variance:2.048+0.576+3.136=5.76Taking the square root of the variance gives the standard deviation:σ=5.76−−−−√≈2.40

Mark receives 10 calls in 3 days on average. How many calls does he receive in 5 days on average?

16.7 calls If μ is the unknown average number of calls over 5 days, then we can set up the proportion: 10 calls3 days=μ5 days Cross multiplying, we find that 3μ=50, so solving for μ, we find thatμ=503≈16.7 calls

Give the numerical value of the parameter p in the following binomial distribution scenario.The probability of winning an arcade game is 0.632 and the probability of losing is 0.368. If you play the arcade game 10 times, we want to know the probability of winning no more than 8 times.Consider winning as a success in the binomial distribution. Do not include p= in your answer.

The probability of winning on an arcade game is 0.659. If you play the arcade game 30 times, what is the probability of winning exactly 21 times? Round your answer to two decimal places.

In this case, we have that x=21, n=30, and p=0.659. Thus, the probability can be found by, P(X=21)=30C21⋅0.65921⋅0.3419=30!21!9!⋅0.65921⋅0.3419≈0.14

Jackie is practicing free throws after basketball practice. She makes a free throw shot with probability 0.7. She takes 20 shots. We say that making a shot is a success. What are p, q, and n in this context?

p=0.7, q=0.3, n=20 Remember that p is the probability of success, which is the probability of making a shot, 0.7. The probability of failure is q=1−p=0.3. n is the number of repetitions, which is 20.

On average, Billy has noticed that 20 trucks pass by his apartment daily (24 hours). In order to find the probability that exactly 4 trucks will pass his apartment in a 3 hour time period using the Poisson distribution, what does the random variable X represent?

the number of trucks in 3 hours The random variable X represents the number of occurrences of an event in the time interval of interest. The time interval of interest is the fixed time period for which the probability of an event is being sought. In this case, the time interval of interest is 3 hours. The random variable X is the number of trucks that pass in the 3 hour time period.

According to a Gallup poll, 60% of American adults prefer saving over spending. Let X= the number of American adults out of a random sample of 50 who prefer saving to spending. What is the mean (μ) and standard deviation (σ) of X?

μ=30 and σ≈3.46 Remember that the mean μ is given by the formula μ=np. This should make sense because you can think of p as the fraction of the sample, on average, that will be a success. In this case p=0.6 because we think of a success as someone who prefers saving over spending. n is the size of the sample, 50. So μ=(50)(0.6)=30 Standard deviation is given by the formula σ=npq−−−√. As above, n=50 and p=0.6. Remember that p+q=1, so solving for q we find that q=1−p=0.4. So σ=(50)(0.6)(0.4)−−−−−−−−−−−√=12−−√≈3.46

Identify the parameters in the following situation. When Jackie is camping, she sees 6 shooting stars every hour, on average. What is the probability that she sees exactly 4 shooting stars in the next hour?

μ=6; x=4; time period is one hour We are asked for the probability of 4 shooting stars in one hour. Therefore, the time period of interest is 1 hour and the value of interest is x=4. The average value μ is given to be 6 shooting stars per hour. So the answer is μ=6; x=4; and the time period is 1 hour.

Alta - Ch. 4

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