AP Stats Probability #51-80

Which of the following statements is incorrect? A. In all normal distributions, the mean and median are equal. B. A normal distribution is completely determined by 2 numbers, its mean adn its standard deviation. C. All bell-shaped curves are normal distributions for some choice of µ and σ. D. Virtually all the area under a normal curve is within 3 standard deviations of the mean, no matter what the particular mean and standard deviation are. E. Standardized scores (z-scores) have a normal distribution only if the original distribution is normal.

E. Because of symmetry the median and mean are always identical for all normal distributions. Many bell-shaped curves are NOT normal curves, and we don't CHOOSE mu and sigma. For all normal curves, 99.7% % of the area is within 3 σ of the mean. When we standardize a variable, the shape of the distribution doesn't change.

Suppose 2 events, E and F, have nonzero probabilities p and q, respectively. Which of the following is impossible? A. p + q > 1 B. p-q < 0 C p/q > 1 D. E and F are neither independent nor mutually exclusive. E. E and F are both independent and mutually exclusive.

E. Independent implies P(E ∩ F) = P(E) x P(F), while mutually exclusive implies P(E ∩ F) = 0. We are given that P(E) and P(F) are nonzero, so it is impossible for E and F to be BOTH independent and mutually exclusive.

Which of the following statements about t-distributions is true? A. The greater the number of degrees of freedom, the narrower the tails. B. The smaller the number of degrees of freedom, the closer the curve is to the normal curve. C. 30 degrees of freedom gives the normal curve. D. The shape of a t-distribution depends on the degrees of freedom which depends on the population size. E. The probability that z>1.96 in a normal distribution is greater than the probability that t>1.96 with df = 30.

A The larger the df, the closer the curve is to the normal curve. Even though around 30 df is often considered a reasonable approximation to the normal curve, it is still usually an approximation and not exactly the same. The degrees of freedom relate to sample size, not population size. For any df, the probability that t>1.96 is greater than the probability that z>1.96 in a normal distribution because tails are fatter in t- than in normal curve.

Among the 125 teachers at a small college, 75 are registered Democrats, 35 are registered Republications and the rest are independents. If a 10-person committee is randomly picked, what is the probability that at least two independents are chosen. A. 10choose2 (¹⁵/₁₂₅)² (¹¹⁰/₁₂₅)⁸ B. 1-(¹⁵/₁₂₅)²+(¹⁵/₁₂₅)¹+(¹⁵/₁₂₅)⁰

C.

Which of the following statements about the chi-square distribution is INCORRECT? A. The probability between two values of chi-square depends on df. B. For small df, the distribution is skewed to the right; however, for large df, it becomes more symmetric and bell-shaped. C. For 1 or 2 degrees of freedom, the histogram peak occurs at 0; for 3 or more degrees of freedom, the peak is at df-2. D. There is a separate chi-square curve for each df value. E. Just like for the t-distribution, the degrees of freedom for chi-square distributions depend upon the sample size.

E. The degrees of freedom for chi-square depend upon numbers of categories, not on sample size. The area under a chi-square is 1, just like any other probability so A is correct. The smaller the df, the more right-skew so B is correct. C is correct. Like t-distribution, there is a separate curve for each df so D is correct.

A new soft drink product has an average number of 77 calories per bottle with a standard deviation of 4.5 calories. In a random sample of 40 bottles, what is the probability that the mean number of calories is between 75 and 80? A. .4191 B. .4975 C. .5000 D. .8383 E. .9975

E. The sampling distribution of x-bar is approx. N with a mean of µ = 77 and a standard deviation of σ = 4.5/√40 = 0.7115. The probability is normalcdf(75, 80, 77, 0.7115) = .9975.

Suppose we have a binomial random variable where the probability of exactly four successes is nC4 p⁴(.37)⁷. What is the mean of the distribution? A. 2.52 B. 2.59 C. 4.07 D. 4.41 E. 6.93

E. This is a binomial with p = 1- .37 = .63, and n = 4+7 = 11. So the mean is np = 11(.63) = 6.93.

Suppose X and Y are random variables with E(X) = 4, E(Y) = 25, and var(Y) = 3. What are the expected value and variance of the random variable X+Y? A. E(X+Y) = 20, var(X+Y) = 7 B. E(X+Y) = 40, var(X+Y) = 3.5 C. E(X+Y) = 40, var(X+Y) = 5 D. E(X+Y) = 40, var(X+Y) = 7 E. There is insufficient information to answer.

E. Without independence we can't assume they can use the rules for adding random discrete variables.

Suppose X and Y are independent random variables, both with normal distributions. If X has mean 30 with standard deviation 6, and Y has mean 25 with standard deviation 4, what is the probability that a randomly generated value of X is greater than a randomly generated value of Y? A. .5000 B. .6914 C. .7440 D. .7560 E. .8413

D. X-Y is normally distributed with mean 30-25 = 5 and standard deviation square root [6²+4²] = 7.211. Then P(X-Y>0) = normalcdf(0, 1000, 5, 7.211) = 0.7560

A. The sampling distribution of p(hat) is approximately normal with mean µ = .7 and standard deviation σ = square root[ (.7)(.3)/84 ] = .05

Which of the histograms represents the sampling distribution of p(hat) for p = .7 and n = 84?

For which of the following is a binomial an appropriate model? A. The number of heads in 10 tosses of an unfair coin weighted so that heads comes up twice as often as tails. B. The number of hits in 5 at-bats where the probability of a hit is either .352 or .324 depending upon whether the pitcher is right-handed or left-handed. C. The number of tosses of a fair coin before heads appears on two consecutive tosses. D. The number of snowy days in a given week. E. The binomial is appropriate in all of the above.

A. Binomial is appropriate for a fixed number of trials and a fixed probability with independent events. In B there is not a fixed probability (p) of success. In C there is not a fixed number of trials. In D the probability of snow on a given day is not independent of whether or not there was snow on a previous day.

Apples growing in a certain orchard have weights that are normally distributed with a standard deviation of 2.2 ounces. What is the mean weight if 80 percent of the apples weigh less than 9.1 ounces? A. 7.25 ounces B. 7.45 ounces C. 7.72 ounces D. 10.95 ounces E. The mean cannot be computed from the information given.

A. Find the critical z-score by computing 80% of the Normal curve, invN(.80, 0, 1) = 0.8416. Then simply use the z-score formula to solve for the mean: z = (actual - mean) / SD 0.8416 = (9.1 - µ ) / 2.2 7.25 = µ

Five managers and 5 employees are on a grievance committee. A three-person subcommittee is formed by a random selection from the 10 committee members. What is the probability that all 3 members of the subcommittee are managers? A. ¹/₁₂ B. ¹/₁₆ C. ³/₁₆ D. ⁵/₁₆ E. ¹/₂

A. There are a total of 10choose3 = 120 ways of picking the subcommittee. Of these, 5choose3 = 10 have all managers. So the probability is 10/120 = 1/12.

Which of the following is a true statement? A. The sampling distribution of the difference x₁-x₂ has a mean equal to the difference of the population means. B. The sampling distribution of the difference x₁-x₂ has a standard deviation equal to the sum of the population standard deviations. C. The sampling distribution for the difference x₁-x₂ has a standard deviation equal to the difference of the population standard deviations. D. The sample sizes must be equal in two-sample inference. E. As long as the sample sizes are large enough (for example, each at least 30), then the samples in two-sample inference do not have to be independent.

A. Variances can be added; standard deviations cannot. The sample sizes n₁ and n₂ do not have to be the same. Without independence, our methods of inference do not apply.

It is known that 66% of the employees at one factory are women, while 57% of the employees of a second factory are women. In an SRS of 75 employees from the first factory and an independent SRS of 60 employees from the second, what is the probability that the difference between the percentages of women picked (first factory minus second) is more than 15%? A. .1486 B. .2378 C. .3064 D. .6189 E. .7622

B. Check that you can use approx normal because of Large Enough condition - both np and nq for the factories are > 10. They are. Then we can model using a sampling distribution of differences p(hat)₁ - p(hat)₂ as Approx Normal with mean .66-.57 = .09 and standard deviation of Square Root[ (p₁q₁)/n₁ + (p₂q₂/n₂) ] = √ (.66)(.34)/75 + (.57)(.43)/60 = .0841. The probability the difference is more than 15% is normalcdf( .15, 1, .09, .0841) = .2378

Given 2 events, E and F, such that P(E) = .340, P(F) = .450 and P(E u F) = .637, then the 2 events are: A. independent and mutually exclusive. B. independent but not mutually exclusive. C. mutually exclusive, but not independent. D. neither independent nor mutually exclusive. E. there is not enough information to answer this question.

B. P(E u F) = P(E) + P(F) = P(E ∩ F), so .637 = .34 + .45 - .153. Since P(E ∩ F) ≠ 0, E and F are not mutually exclusive. However, P(E ∩ F) = .153 (.34 x .45), which implies that E and F are independent.

Suppose that 62% of the graduates from your high school go on to four-year colleges, 15% go on to two-year colleges, 18% find employment, and the remaining graduates search for a job. If a randomly selected student is not going on to a four-year college, what is the probability he or she will find employment? A. .440 B. .474 C. .526 D. .545 E. .560

B. The probability someone will be searching for a job is 1 - (.62+.15+.18) = .05. then .18/(.15 +.18+.05) = .474. OR .18/(1-.62) = .474.

It is known that shoppers at one department store show a mean satisfaction index of 8.3 with a standard deviation of 0.4 on a standardized test, while shoppers at a second store show a mean index of 7.9 with a standard deviation of 0.3 on the same test. In independent SRSs of 100 customers each, what is the probability that the average index in the sample from the first store is more than a half point higher than the average from the second store? A. .0114 B. .0228 C. .0455 D. .0766 E. .4432

B. The sampling distribution of x₁-x₂ is approximately normal with a mean of 8.3-7.9 = 0.4 and a standard deviation of = square root[ (0.4)²/100 + (0.3)²/100 = 0.05 The probability that the difference is more than 0.5 is normalcdf(0.5, 100, 0.4, 0.05) = .0228.

Which of the following statements is true? A. The mean of the set of sample means varies inversely with the square root of the size of the sample. B. The variance of the set of sample means varies directly with the size of the samples and inversely with the variance of the original population. C. The standard deviation of the set of sample means varies directly with the standard deviation of the original population and inversely with the square root of the size of the samples. D. The larger the sample size, the larger the variance of the set of sample means. E. One must double the sample size in order to cut the standard deviation of x-bar in half.

C. A is wrong because the mean of the set of sample means is equal to the mean of the population: it does not vary with the size of the samples! B is wrong because variance of the set of sample means varies inversely with the size of the samples and directly with the variance of the original population sample σ = population σ/√n. E is wrong because you must take a sample FOUR times as large in order to cut the standard deviation of x-bar in half.

Following are parts of the probability distributions for the random variables X and Y. X | P(X) 1 | ? 2 | .2 3 | .3 4 | ? Y | P(Y) 1 | .4 2 | ? 3 | .1 If X and Y are independent and the joint probability P(X=1,Y=2) = 0.1, what is P(X=4)? A. .1 B. .2 C. .3 4. .4 E .5

C. First find P(Y=2). P(Y=2) = 1 - minus all others (.4 and .1) P(Y = 2) = .5 By independence, P(X=1,Y=2) = P(X=1) times P(Y=2), and so substitute and solve: .1 = P(X=1)(.5) Divide both sides by .5 and find P(X=1) = .2 Next, P(X=4) = 1 - all others (.2 + .2 + .3) P(X=4) = .3

Suppose that for a certain coastal city, in any given year the probability of a major hurricane hitting is .4, the probability of flooding is .3, and the probability of both a major hurricane and flooding is .2. What is the probability of flooding if a major hurricane hits? A. .200 B. .286 C. .500 D. .667 E. .750

C. P(flooding | hurricane) = P(flooding ∩ hurricane) / P(hurricane) = .2/.4 = .5

The number of hybrid cars a dealer sells weekly has the following probability distribution: The dealer purchases the cars for $21,00 and sells them for 424,500. What is the expected weekly profit from selling hybrid cars? A. $2,380 B. $3,500 C. $5,355 D. $8,109 E. $37, 485

C. The expected number of cars sold per week is 0(.32) + 1(.28) + 2(.15) + 3(.11) + 4(.08) + 5(.06) = 1.53. With a profit for each car of $24,500-$21,000 = #3,500, the expected weekly profit is found by multiplying 1.53 x 3500 = $5,355.

Three fair coins are tossed. If all land "heads," the player wins $10, and if exactly two land heads, the player wins $5. If it costs $4 to play, what is the player's expected outcome after four games? A. Loss of $0.875 B. Loss of $1.00 C. Loss of $3.50 D. Win of $2.25 E Win of $9.00

C. The expected winnings in one game is found by computing E(X), expected value. = 10(.5)³ + 5(3(.5)²(.5) - 4 = -8.75 For 4 games, multiply by 4 4(-8.75) = -$3.50

The sampling distribution of the sample mean is close to the normal distribution A. only if the parent population is unimodal, not badly skewed, and does not have outliers. B. no matter what the distribution of the parent population of what the value of n. C. if n is large, no matter what the distribution of the parent population. D. if the standard deviation of the parent population is known. E. only if both n is large and parent population has a normal distribution.

C. This follows from the Central Limit Theorem.

Suppose a manufacturer knows that 20% of the circuit boards coming off the assembly line have a minor defect. If an inspector keeps inspecting boards until he comes upon one with the defect, what is the probability he will have to inspect at most three boards? A. .128 B. .384 C. .488 D. .512 E. .896

C. This is a cumulative geometric setting with probability .20 + (.80)(.20) + (.80)²(.20) = .488

The mean income per household in a certain state is $28,000 with a standard deviation of $8,500. Assuming a normal distribution, 95% of all households have an income over what amount? A. $11,000 B. $11,300 C. $14,000 D. $19,500 E. $42,000

C. invNorm( .05, 28000, 8500) = $14,000. * Use .05 because it is the left 5% of the tail, filling left to right.

In a set of 8 boxes, 3 boxes each contain 2 red and 2 green marbles, while the remaining boxes each contain 3 red and 2 green marbles. A player randomly picks a box and then randomly picks a marble from that box. She wins if she ends up with a red marble. If she plays four times, what is the probability she wins exactly twice? A. .0606 B. .3164 C. .3221 D. .3634 E. .5625

D. The probability of winning (ending up with a red marble) is ³/₈ x ¹/₂ + ⁵/₈ x ³/₅ = ⁹/₁₆. Using this, the probability of winning exactly twice in four games is expressed as winning twice (⁹/₁₆) times losing the other two games (the difference which is...) (⁷/₁₆). (⁹/₁₆)(⁷/₁₆) = .3634. The first step is best found by simply using a Conditional Tree Diagram.

The SAT math scores for applicants to a particular engineering school are normally distributed with a mean of 680 and a standard deviation of 35. Suppose that only applicants with scores above 700 are considered for admission. What percentage of the applicants considered have scores below 750? A. 2.3% B. 28.4% C. 71.6% D. 92.0% E. 97.7%

D. The probability that an applicant is considered is normalcdf (700, 1000, 680, 35) = .2339. The probability that an applicant has a score above 750 is normalcdf (750, 1000, 680, 35) = .0228. The percentage below 750 given that the scores are above 700 is (.2839 - .0228) / .2839 = 92.0%.

The distribution of weights of '16oz' potato chip bags is given by the histogram image here. The distribution has a mean of 16.96 oz with a standard deviation of 2.31 oz. If 50 random samples of 12 bags each are picked, and the mean weight of each sample is found, which of the following is most likely to represent the distribution of all the sample means?

D. The sampling distribution for x-bar should be roughly bell-shaped with mean around 17 and standard deviation about 2.31/√12 = 0.67. Choice D is the most reasonable.