SOLUTIONS TO MIDTERM
VERSION 1
1) Let X = profit on a given play. The two possible values for X are .67 (if the listener buys the song) and −.03 (if the listener does not buy the song). We have E[X] = .67 p − .03 (1−p) = .7p − .03. Setting E[X] = 0 and solving for p yields p = .03 /.7 = .0429. Answer is D.
2) We use the formula.
This formula implies that as the sample size n is increased, since the population standard deviation σ is fixed, the standard error of the mean approaches zero. (The sample mean becomes a more accurate estimator of the population mean μ as the sample size is increased.) Answer is C.
3) Let X be the number who support Hillary Clinton. Then X has a binomial distribution with n=3, p=.23. We need
Answer is C.
4) Let X be the number of supernovas observed next year. We have that X is normally distributed with standard deviation σ and mean μ = 3σ. We have .1587 = Prob{X≥1} = Prob{Z≥(1−μ)/σ} = Prob{Z ≥ (1−3σ)/σ} =Prob{Z≥1/σ −3}. Note that .1587 = .5 − .3413 is the probability that a standard normal exceeds 1. Therefore, we have1/σ − 3 = 1. It follows that σ = 1/4. We know that μ = 3σ = 3/4 = .75. Answer is B.
5) Let A = {Adele will release an album next year}, B = {Adele will go on tour next year}. We have P(B|A) = .6, P(A)=.3. Note that P(A)= P(AÇB)+ P(A Ç ) . We need to calculate P(A Ç ) =P(A)−P(AÇB)=P(A)−P(A)P(B|A)=.3−(.3)(.6)=.12. Answer is A.
6) Prob{Z≥−.35}=Prob{−.35≤Z≤0}+Prob{0≤Z < ∞}=.1368+.5=.6368. Answer is B.
7) Converting to z-scores, we need Prob{Z < (5−3)/2} = Prob {Z < 1} = .5 + .3413 = .8413. Answer is D.
8) As pointed out in Handout 9 (sampling), the formula says that the sample mean is unbiased in estimating the population mean. Answer is C.
9) We have Answer is E.
10) First, we compute E[X] = 1 (.2) + (−3)(.8) = −2.2. Next, we get Var(X) = (1+2.2)2 (.2) + (−3+2.2)2 (.8) = 2.56.
Therefore, SD(X) = = 1.6. Answer is D.
11) Note that A Ç B = {Total is 11} = A. So A, B are not mutually exclusive. Of the 36 equally likely outcomes for the two dice, two of them give a total of 11, so P(A) = 2/36. The outcomes that lead to a total of at least 10 are: (6,4),(5,5),(4,6),(6,5),(5,6),(6,6), so P(B)=6/36. Since 2/36 is not equal to (2/36) (6/36), we have that P(A Ç B) ≠ P(A) P(B), so A, B are not independent. Answer is C. Another way to see that A, B are not independent is to notice that P(B|A) = 1, which is not equal to P(B). Answer is C.
12) Use the Central Limit Theorem. We have = 130000, and = 40000 / 5 = 8000. Then using the normal distribution, we get Prob{130000 < < 140000} = Prob{(130000−130000)/8000 < Z < (140000−130000)/8000}=Prob{0<Z<1.25} = .3944 from Table 5.
Answer is D.
13) We have E[X] = np = 4/3, and The z-score corresponding to x=1 is
Answer is A.
14) With the original (smaller) sample size, the standard error is . With the larger sample size, the standard error is . Taking the ratio of the new to the old standard error yields
Answer is D.
15) Since the sample mean is zero, the two data points sum to zero. Let's label the two data points as x and −x (where x>0). Since the degrees of freedom is 1, the sample variance is Since the sample standard deviation is , the sample variance is 8, and we get , so so The data points are 2 and -2. The smallest
data point is −2, corresponding to a z-score
Answer is B.