Chapter 5 - Probability Distributions
TYPES OF RANDOM VARIABLES
- A random variable is discrete if the number of possible outcomes is countable (Integers) and can be plotted on a number line with space between each point. - A random variable is continuous if it can take on any value within an interval (not just integers, but any value in between) The possible outcomes cannot be listed individually. -- Continuous random variables can be plotted over an uninterrupted region of the number line
random variable
- A random variable, x, is a numerical outcome from a probability experiment, whose value is determined by chance. x = The number of people in a randomly chosen car x = The time you have to wait for an elevator to arrive x = The number of successful basketball shots, in 5 tries.
The expected value is 2.438
- Most of the household sizes differ from the expected value by 1 or 2 people. - The average household is expected to have either 2 or 3 people.
MEAN of a Discrete Random Variable
- The mean of a discrete probability random variable would be the balance point of the probability histogram. - The mean (denoted μ ) predicts the average of the outcomes, if the experiment were repeated over and over. - Mean = E(x) = μ = sigma x * P(x)
probability distribution
- the possible values of the random variable, - along with their corresponding probabilities. - A probability distribution can be in the form of a table, graph or mathematical formula.
An investment counselor calls with a hot stock tip. He believes that if the economy remains strong, the investment will result in a profit of $50,000. If the economy grows at a moderate pace, the investment will result in a profit of $10,000. However, if the economy goes into recession, the investment will result in a loss of $50,000. You contact an economist who believes there is a 30% probability the economy will remain strong, a 60% probability the economy will grow at a moderate pace, and a 10% probability the economy will slip into recession. What is the expected profit from this investment?
.3*50000+.6*10000-.10*50000 = 16000
Determine the required value of the missing probability to make the distribution a discrete probability distribution. x P(x) 3 0.29 4 ? 5 0.39 6 0.09
1-0.29-0.39-0.09 = 0.23
A probability distribution MUST satisfy the following rules:
1. Each probability must be between 0 and 1 (inclusive) [0 <= P(x) <= 1] 2. The sum of all probabilities must equal 1.
The four conditions of a binomial experiment:
1. The experiment is repeated a fixed number of trials. 2. The trials are independent. 3. Each trial has 2 outcomes: Success or Failure 4. The probability of success, p, stays the same on every trial.
What is a discrete probability distribution? What are the two conditions that determine a probability distribution?
A discrete probability distribution lists each possible value a random variable can assume, together with its probability. The probability of each value of the discrete random variable is between 0 and 1, inclusive, and the sum of all the probabilities is 1.
The number of baskets you MAKE in 5 shots
Discrete: x = 0, 1, 2, 3, 4, 5
The number of people in a randomly chosen car
Discrete: x = 0, 1, 2, 3, 4, 5
The number of light bulbs that burn out in a room of 10 light bulbs in the next year.
Discrete: x =0, 1, 2, 3....10 burned out bulbs
Is the distance a baseball travels in the air after being hit a discrete random variable, a continuous random variable, or not a random variable?
It is a continuous random variable.
Is the height of a randomly selected giraffe a discrete random variable, a continuous random variable, or not a random variable?
It is a continuous random variable.
Is the square footage of a pool/house a discrete random variable, continuous random variable, or not a random variable?
It is a continuous random variable.
Is the number of free dash throw attempts before the first shot is made a discrete random variable, a continuous random variable, or not a random variable?
It is a discrete random variable
Is the number of people in a restaurant that has a capacity of 300 a discrete random variable, a continuous random variable, or not a random variable?
It is a discrete random variable
Is the number of statistics students now doing their homework a discrete random variable, continuous random variable, or not a random variable?
It is a discrete random variable.
"USUAL":
Most values lie within 2 standard deviations of the mean.
Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. Seven cards are selected from a standard 52-card deck without replacement. The number of sevens selected is recorded.
No, because the trials of the experiment are not independent and the probability of success differs from trial to trial.
Seven cards are selected from a standard 52-card deck without replacement. The number of nines selected is recorded.
No, because the trials of the experiment are not independent and the probability of success differs from trial to trial.
Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Recording the number of televisions in 150 households nothing
No, because there are more than two possible outcomes
Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Randomly selecting 50 citizens of a state and recording their nationalities nothing
No, because there are more than two possible outcomes
Determine whether or not the procedure described below results in a binomial distribution. If it is not binomial, identify at least one requirement that is not satisfied. Five hundred different voters in a region with two major political parties, A and B, are randomly selected from the population of 4000 registered voters. Each is asked if he or she is a member of political party A, recording Yes or No.
No, the trials are not independent and the sample is more than 5% of the population
d. If at most one household is tuned to Lindsay and Tobias, does it appear that the 20% share value is wrong? Why or why not?
No, because with a 20% rate, the probability of at most one household is greater than 0.05.
Assume that a procedure yields a binomial distribution with n=5 trials and a probability of success of p=0.30. Use a binomial probability table to find the probability that the number of successes x is exactly 4.
P(4) = .028
The table to the right lists probabilities for the corresponding numbers of girls in three births. What is the random variable, what are its possible values, and are its values numerical? Number_of_girls P(x) 0 0.125 1 0.375 2 0.375 3 0.125
The random variable is x, which is the number of girls in three births. The possible values of x are 0, 1, 2, and 3. The values of the random value x are numerical.
Which of the following is not a requirement of the binomial probability distribution?
The trials must be dependent.
Which of the following is NOT one of the three methods for finding binomial probabilities that is found in the chapter on discrete probability distributions?
Use a simulation
Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Surveying 150 college students and asking if they like pirates or ninjas better comma recording Yes or No
Yes, because all 4 requirements are satisfied
Randomly selecting 200 adults and asking if they like rap music comma recording Yes or No
Yes, because all 4 requirements are satisfied
Treating 50 men with a special shampoo and recording Yes if they experience any burning or No otherwise
Yes, because all 4 requirements are satisfied
Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Surveying 50 teenagers and recording if they have ever committed a crime nothing
Yes, because all four requirements are met
Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 130 randomly selected individuals, with the number of individuals responding favorably recorded.
Yes, because the experiment satisfies all the criteria for a binomial experiment.
Six hundred different voters in a region with two major political parties, A and B, are randomly selected from the population of 2.9 million registered voters. Each is asked if he or she is a member of political party A, recording Yes or No.
Yes, the result is a binomial probability distribution
One of the cupholders in your car has leftover coins in it. The cupholder has 3 pennies and 5 nickels, and you reach in and pick TWO without replacement. a) If we define the random variable, x, to equal the total value of the two coins you are holding in your hand, what is the highest amount you could have? b) If we define the random variable, x, to equal the total value of the two coins you are holding in your hand, what is the lowest amount you could have? c) Are there any other amounts possible, with the two coins you are holding?
a) 10 cents b) 2 cents c) yes, the two coins could also add up to 6 cents d) x: 2, 6, 10 P(x): .107, .536, .357
Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The number of statistics students now reading a book b. The time it takes for a light bulb to burn out c. The gender of college students d. The number of hits to a website in a day e. The exact time it takes to evaluate 27 plus 72 f. The time it takes to fly from City Upper A to City Upper B
a. It is a discrete random variable. b. It is a continuous random variable. c. It is not a random variable d. It is a discrete random variable. e. It is a continuous random variable f. It is a continuous random variable
Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The number of people with blood type Upper A in a random sample of 14 people b. The number of fish caught during a fishing tournament c. The response to the survey question "Did you smoke in the last week question mark " d. The height of a randomly selected giraffe e. The number of light bulbs that burn out in the next week in a room with 10 bulbs f. The square footage of a house
a. It is a discrete random variable. b. It is a discrete random variable. c. It is not a random variable. d. It is a continuous random variable. e. It is a discrete random variable. f. It is a continuous random variable.
In a region, there is a 0.9 probability chance that a randomly selected person of the population has brown eyes. Assume 10 people are randomly selected. Complete parts (a) through (d) below. a. Find the probability that all of the selected people have brown eyes. b. Find the probability that exactly 9 of the selected people have brown eyes. c. Find the probability that the number of selected people that have brown eyes is 8 or more. d. If 10 people are randomly selected, is it unusual for 8 or more to have brown eyes?
a. P(10) = .349 b. P(9) = .387 c. P(8 or 9 or 10) = .194 + .387 + .349 = .93 d. The probability that the number of people with brown eyes is 8 or more, P(8 or 9 or 10) is greater than the cutoff of 0.05. No
binomial random variable
the number of successes when something is repeated n times.
Groups of people aged 15-65 are randomly selected and arranged in groups of six. In the accompanying table, the random variable x is the number in the group who say that their family and/or partner contribute most to their happiness (based on a survey). x P(x) 0 0+ 1 0.025 2 0.028 3 0.104 4 0.271 5 0.449 6 0.123
x P(x) x*P(x) (x - μ)^2*P(x) 0 0 0.00 0.00 1 0.021 0.021 0.202 2 0.027 0.054 0.119 3 0.260 0.780 0.315 4 0.257 1.028 0.003 5 0.345 1.725 0.279 6 0.090 0.540 0.325 Total μ = 4.1 σ^2 = 1.243 σ = 1.1
The _______ of a discrete random variable represents the mean value of the outcomes.
expected value
Unusually low:
if P(x or fewer) <= 0.05
Unusually high:
if P(x or more) <= 0.05
P(x)
is the probability of getting exactly x successes among the n trials.
A candy company claims that 19% of its plain candies are orange, and a sample of 200 such candies is randomly selected. b. A random sample of 200 candies contains 59 orange candies. Is this result unusual? Does it seem that the claimed rate of 19% is wrong?
mean = 38 standard deviation = 5.5 Yes, because 59 is greater than the maximum usual value. Thus, the claimed rate of 19% is probably wrong.
If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning. a. Find the mean and standard deviation for the number of wins of gamblers who bet on the number 7 one hundred and seventy times. b. Would 0 wins in one hundred and seventy bets be an unusually low number of wins?
n = 170 p = 0.0263 q = 0.9737 find mean and standard dev
In a clinical trial of a drug used to help subjects stop smoking, 832 subjects were treated with 1 mg doses of the drug. That group consisted of 33 subjects who experienced nausea. The probability of nausea for subjects not receiving the treatment was 0.0114. Complete parts (a) through (c). a. Assuming that the drug has no effect, so that the probability of nausea was 0.0114, find the mean and standard deviation for the numbers of people in groups of 832 that can be expected to experience nausea. b. Is it unusual to find that among 832 people, there are 33 who experience nausea? c. Based on the preceding results, does nausea appear to be an adverse reaction that should be of concern to those who use the drug?
n = 832 p = .0114 μ = np = 9.5 σ = 3.1 μ-2σ = 3.3 μ+2σ = 15.7 It is unusual because 33 is outside the range of usual values. The drug does appear to be the cause of some nausea. Since the nausea rate is still quite low (about 4%), it appears to be an adverse reaction that does not occur very often.
In the binomial probability formula, the variable x represents the _______.
number of successes.
Discrete
random variables are counted
Continuous
random variables are measured
n
stands for the fixed number of trials.
x
stands for the number of successes (in n trials, x can be any whole number between 0 and n)
q
stands for the probability of failure in one trial.
p
stands for the probability of success in one trial.
Ted is not particularly creative. He uses the pickup line "If I could rearrange the alphabet, I'd put U and I together." The random variable x is the number of girls Ted approaches before encountering one who reacts positively. Determine whether the table describes a probability distribution. If it does, find its mean and standard deviation. x P(x) 1 0.002 2 0.014 3 0.111 4 0.274 5 0.353
sum[P(x)] = .754 since doesn't equal 1, table is not a probability distribution
The length of time between calls to 911.
Continuous: x >= 0
The time you have to wait for an elevator to arrive
Continuous: x >= 0
The gallons of gas bought by randomly chosen customer at a gas station
Continuous: x >= 0
A headline in a certain newspaper states that "most stay at first job less than 2 years." That headline is based on an online poll of 380 college graduates. Among those polled, 80% stayed at their first full-time job less than 2 years. Complete parts (a) through (d). a. Assuming that 50% is the true percentage of graduates who stay at their first job less than two years, find the mean and standard deviation of the numbers of such graduates in randomly selected groups of 380 graduates. b. Assuming that the 50% rate in part (a) is correct, find the range of usual values for the numbers of graduates among 380 who stay at their first job less than two years. c. Find the actual number of surveyed graduates who stayed at their first job less than two years. Use the range of values from part (b) to determine whether that number is unusual. Does the result suggest that the headline is not justified? d. The statement, "Alumni who opted-in to receive communications from the website were invited to participate in the online poll, and 380 of them completed the survey," was given as part of the description of the survey methods used. What does that statement suggest about the results?
a. n = 380 p = 0.5 mean = 190 standard deviation = 9.7 b. max = 170.6 min = 209.4 c. .8 .8*380 = 304 Since 304 is outside of the range of usual values found in part (b), it is unusual. d. voluntary response sample
In a past election, the voter turnout was 58%. In a survey, 1071 subjects were asked if they voted in the election. a. Find the mean and standard deviation for the numbers of voters in groups of 1071. b. In the survey of 1071 people, 627 said that they voted in the election. Is this result consistent with the turnout, or is this result unlikely to occur with a turnout of 58%? Why or why not? c. Based on these results, does it appear that accurate voting results can be obtained by asking voters how they acted?
a. μ = 621.2 σ = 16.2 b. This result is usual because 627 is within the range of usual values c. Yes, because the results indicate that 58% is a possible turnout
For a multistate lottery, the following probability distribution represents the cash prizes of the lottery with their corresponding probabilities. x (cash prize, $) P(x) 12,000,000 0.00000000804 200,000 0.00000026 10,000 0.000001865 100 0.000142603 7 0.003340745 4 0.007541141 3 0.01259363 0 0.97637974796 (a) If the grand prize is $12,000,000, find and interpret the expected cash prize. If a ticket costs $1, what is your expected profit from one ticket? The expected cash prize is (b) What is the correct interpretation of the expected cash prize? (c) The expected profit from one $1 ticket is
a. $0.27 b. On average, you will win $0.27 per lottery ticket. c. -$.73
Use the probability distribution to find probabilities in parts (a) through (c). Dogs Households 0 0.676 1 0.203 2 0.071 3 0.024 4 0.017 5 0.009 (a) Find the probability of randomly selecting a household that has fewer than two dogs. (b) Find the probability of randomly selecting a household that has at least one dog. (c) Find the probability of randomly selecting a household that has between one and three dogs, inclusive.
a. .676+.203 = .879 b. .203+.071+.024+.009 = .324 c. .203+.071+.024 = .298
The accompanying table describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods. Complete parts (a) through (d). x (Number of Peas with Green Pods) P(x) 0 0+ 1 0+ 2 0.004 3 0.022 4 0.083 5 0.176 6 0.312 7 0.298 8 0.105 a. Find the probability of getting exactly 7 peas with green pods. b. Find the probability of getting 7 or more peas with green pods. c. Which probability is relevant for determining whether 7 is an unusually high number of peas with green pods, the result from part (a) or part (b)? d. Is 7 an unusually high number of peas with green pods? Why or why not? Use 0.05 as the threshold for an unusual event.
a. 0.298 b. P(7 or more) = 0.298 + 0.105 = 0.403 c. P(7 or more) d. Since P(7 or more) is greater than 0.05, it is not an unusually high number
Based on data from a car bumper sticker study, when a car is randomly selected, the number of bumper stickers and the corresponding probabilities are as shown below. x P(x) 0 0.776 1 0.101 2 0.048 3 0.023 4 0.015 5 0.013 6 0.009 7 0.007 8 0.006 9 0.002 a. Does the given information describe a probability distribution? b. Assuming that a probability distribution is described, find its mean and standard deviation. c. Use the range rule of thumb to identify the range of values for usual numbers of bumper stickers. d. Is it unusual for a car to have more than one bumper sticker? Explain.
a. 0.776 + 0.101 + 0.048 + 0.023 + 0.015 + 0.013 + 0.009 + 0.007 + 0.006 + 0.002 = 1 Yes b. x P(x) x*P(x) (x - μ)^2*P(x) 0 0.776 0.000 0.279 1 0.101 0.101 0.016 2 0.048 0.096 0.094 3 0.023 0.069 0.132 4 0.015 0.060 0.173 5 0.013 0.065 0.252 6 0.009 0.054 0.262 7 0.007 0.049 0.287 8 0.006 0.048 0.329 9 0.002 0.018 0.141 Total μ = 0.6 σ^2 = 1.965 μ = 0.6 σ = 1.4 c. maximum usual value = μ + 2σ = 0.6+2*1.4 = 3.4 minimum usual value = μ - 2σ = 0.6 -2*1.4 = -2.2 (put 0 as min) d. P(more than 1) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) = 0.048 + 0.023 + 0.015 + 0.013 + 0.009 + 0.007 + 0.006 + 0.002 = 0.123 Since the probability that a car has more than one bumper sticker is 0.123, which is greater than 0.05, it is not unusual for a car to have more than one bumper sticker.
A certain TV show recently had a share of 85, meaning that among the TV sets in use, 85% were tuned to that show. Assume that an advertiser wants to verify that 85% share value by conducting its own survey, and a pilot survey begins with 9 households having TV sets in use at the time of the TV show broadcast. Complete parts (a) through (d) below. a. Find the probability that all of the households are tuned to the TV show. b. Find the probability that exactly 8 households are tuned to the TV show. c. Find the probability that at least 8 households are tuned to the TV show. d. If at least 8 households are tuned to the TV show, does it appear that the 85% share value is wrong? Why or why not?
a. 0.85^9 = .232 b. P(8) = .368 c. P(8) + P(9) = .232 + .368 = .600 d. No, because 8 households tuned to the TV show is not unusually high if the share is 85%.
A frequency distribution is shown below. Complete parts (a) and (b). Televisions Households 0 29 1 450 2 723 3 1409 (a) Use the frequency distribution to construct a probability distribution. (b) Graph. Describe the histogram's shape. Choose the correct answer below.
a. 29+450+723+1409 = 2611 P(x) = households/total 29/2611, 450/2611, 723/2611, 1409/2611 b. skewed left
Researchers conducted a study to determine whether there were significant differences in graduation rates between medical students admitted through special programs and medical students admitted through the regular admissions criteria. It was found that the graduation rate was 94% for the medical students admitted through special programs in all medical schools. Complete parts (a) and (b) below. a. If 14 of the students admitted through special programs are randomly selected, find the probability that at least 13 of them graduated. b. Suppose a medical school has 14 students admitted through one of the special programs of its medical program. Does the probability calculated in part (a) apply to these students?
a. P(13 or 14) n = 14 p = 0.94 P(13) + P(14) = .376 + .421 = .797 b. No, because the students admitted through a single special program at a specific medical school are not a random sample.
Assume that 12 jurors are selected from a population in which 50% of the people are Mexican-Americans. The random variable x is the number of Mexican-Americans on the jury. x P(x) 0 0.000 1 0.003 2 0.016 3 0.054 4 0.121 5 0.193 6 0.226 7 0.193 8 0.121 9 0.054 10 0.016 11 0.003 12 0.000 a. Find the probability of exactly 2 Mexican-Americans among 12 jurors. b. Find the probability of 2 or fewer Mexican-Americans among 12 jurors. c. Which probability is relevant for determining whether 2 jurors among 12 is unusually low: the result from part (a) or part (b)? d. Is 2 an unusually low number of Mexican-Americans among 12 jurors? Why or why not?
a. P(2) = .016 b. 0.000 + 0.003 + 0.016 = 0.019 c. The result from part (b), because it measures the probability of 2 or fewer successes. d. It is unusual to have only 2 Mexican-American jurors out of 12 because the probability is less than or equal to 0.05.
An airline has a policy of booking as many as 23 persons on an airplane that can seat only 22. (Past studies have revealed that only 91.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 23 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur? a. The probability that not enough seats will be available is: ___________ b. Is it unlikely for such an overbooking to occur?
a. P(23) n = 23 p = .91 P(23) = .1142 b. It is not unlikely for such an overbooking to occur, because the probability of the overbooking is greater than 0.05.
Multiple-choice questions each have five possible answers left parenthesis a comma b comma c comma d comma e right parenthesis, one of which is correct. Assume that you guess the answers to three such questions. a. Use the multiplication rule to find P(WCW), where C denotes a correct answer and W denotes a wrong answer b. Beginning with WCW, make a complete list of the different possible arrangements of one correct answer and two wrong answers, then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made?
a. P(C) = 1/5 P(W) = 4/5 P(WCW) = P(W) * P(C) * P(W) = (4/5)(1/5)(4/5) = 0.128 b. P(CWW) = 0.128 P(WWC) = 0.128 c. P(WCW) + P(CWW) + P(WWC) = 0.128 + 0.128 + 0.128 = 0.384
Seven peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 7 offspring peas, no more than 1 has a green pod. Is it unusual to get no more than 1 pea with a green pod when 7 offspring peas are generated? Why or why not? a. The probability that no more than 1 of the 7 offspring peas has a green pod is nothing: ________ b. Is it unusual to randomly select 15 peas and find that no more than 1 of them has a green pod? Note that a small probability is one that is less than 0.05.
a. P(no more than 1) = P(1) + P(0) n = 7; x = 1; p = 0.75; q = 1-p = 1-0.75 = 0.25 P(1) = 7!/(7-1)!1! * 0.75^1 * 0.25^(7-1) = 0.001 P(0) = 7!/(7-0)!0! * 0.75^0 * 0.25^(7-0) = 0.000 P(no more than 1) = P(1) + P(0) = 0.001 + 0 = 0.001 b. Yes, because the probability of this occurring is very small
A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 19 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 5% rate of defects, what is the probability that this whole shipment will be accepted? a. The probability that this whole shipment will be accepted is _______
a. P(X=0) + P(X=1) n = 19 p = 0.05 q = 0.95 P(0) + P(1) = 0.755
Expected Value
can be used to predict long run profit or loss.
The accompanying data table describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children. Complete the questions below. x P(x) 0 0.004 1 0.011 2 0.035 3 0.117 4 0.206 5 0.239 6 0.205 7 0.114 8 0.041 9 0.019 10 0.009 Use the range rule of thumb to identify a range of values containing the usual numbers of girls in 10 births.
x P(x) x*P(x) (x - μ)^2*P(x) 0 0.005 0.000 0.130 1 0.010 0.010 0.168 2 0.046 0.092 0.442 3 0.113 0.339 0.498 4 0.194 0.776 0.235 5 0.241 1.205 0.002 6 0.211 1.266 0.171 7 0.111 0.777 0.401 8 0.039 0.312 0.328 9 0.020 0.180 0.304 10 0.010 0.100 0.240 Total μ = 5.1 σ^2 = 2.919 σ = 1.7 maximum usual value = μ + 2σ = 5.1 + 2*1.7 = 8.5 minimum usual value = μ - 2σ = 5.1 -2*1.7 = 1.7 Yes, 1 girl is an unusually low number of girls, because 1 girl is outside of the range of usual values.
In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. x P(x) 0 0.03 1 0.11 2 0.32 3 0.24 4 0.16 5 0.14
x P(x) x*P(x) (x-μ)^2*P(x) 0 0.03 0 0.16 1 0.11 .09 0.29 2 0.32 .64 0.20 3 0.24 1.02 0.01 4 0.16 .56 0.20 5 0.14 .45 0.44 μ = sum[x*P(x)] = 2.8 σ^2 = sum[(x-μ)^2*P(x)] = 1.3 σ = 1.1
The accompanying table describes results of road worthiness tests of cars that are 3 years old. The random variable x represents the numbers of cars that failed among six that were tested for road worthiness. Find the mean and standard deviation for the number of cars that failed among the six cars that are tested. x_(Number_of_cars_that_failed) P(x) 0 0.271 1 0.297 2 0.243 3 0.091 4 0.098 5 0+ 6 0+
x*P(x) [(x-μ)^2*P(x)] 0 .543 .296 .047 .484 .087 .333 .284 .296 .500 0 0 0 0 μ = 1.4 σ^2 = 1.461 σ = 1.2
What are the two requirements for a discrete probability distribution?
ΣP(x) = 1 0 ≤ P(x) ≤ 1
Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.49. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls? x P(x) 0 0.133 1 0.382 2 0.367 3 0.118
μ = 0*0.216 + 1*0.432 + 2*0.288 + 3*0.064 = 1.2 σ = sqrt [sum[(x - μ)^2*P(x)] - μ^2] = sqrt[0^2*.216 + 1^2 + .432 + 2^2*.288 + 3^2*0.064 - 1.2^2] = 0.85
Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. x P(x) 0 0.2707 1 0.4043 2 0.2415 3 0.0721 4 0.0108 5 0.0006
μ = 1(.4043) +2(.2415)+3(.0721)+(4*0.0108)+5(0.0006) = 1.1 x P(x) x^2*P(x) 0 0.2707 0 1 0.4043 .4043 2 0.2415 .966 3 0.0721 .6489 4 0.0108 .1728 5 0.0006 .015 sum[x^2*P(x)] = 2.207 σ^2 = 2.2070-(1.1)^2 = .997 σ = 1.0
Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ-2σ and the maximum usual value μ+2σ. n = 1425, p = 4/5
μ = np = 1425(4/5) = 1140 q = 1-p = 1-4/5 = 1/5 σ = sqrt(npq) = 15.1 μ-2σ = 1140-2(15.1) = 1109.8 μ+2σ = 1140+2(15.1) = 1170.2
Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ-2σ and the maximum usual value μ+2σ. In an analysis of preliminary test results from a gender-selection method, 19 babies are born and it is assumed that 50% of babies are girls, so n=19 and p=0.5.
μ = np = 19*0.5 = 9.5 q = 1-p = 11-0.5 = 0.5 σ = sqrt(npq) = 2.2 μ-2σ = 9.5-2(2.2) = 5.1 μ+2σ = 9.5+2(2.2) = 13.9
Is the expected value of the probability distribution of a random variable always one of the possible values of x? Explain.
No, because the expected value may not be a possible value of x for one trial, but it represents the average value of x over a large number of trials.
An experimental drug is administered to 80 randomly selected individuals, with the number of individuals responding favorably recorded.
Yes, because the experiment satisfies all the criteria for a binomial experiment.
Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 60 randomly selected individuals, with the number of individuals responding favorably recorded.
Yes, because the experiment satisfies all the criteria for a binomial experiment.
