You may use a crib sheet and a calculator, but no books or notes, on this examination. The point value of each problem is shown in brackets. There are 30 points. Express all answers to the nearest thousandth. Indicate clearly what you are doing at each step. This will maximize your chances for receiving partial credit if your final answer is wrong.
1. Here is a random sample of eleven scores on a Math 320 exam:
a) [1]. Find the sample mean.
Answer: xbar = 21.545 = 21.5, by calculator.
b) [1]. Find the sample standard deviation.
Answer: sx = 5.24 by calculator (population s.d. 4.998 is incorrect answer).
c) [1]. Find the median.
Answer: 23
d) [1]. Find the $z$-score for the student who received the highest score on the exam.
Answer: z = (29 - xbar)/sx = 1.42
Here is an edf of this data.
e) [1]. Find the interquartile range.
Answer: From the edf, since 2/11 < .25 < 3/11 and 8/11 < .75 < 9/11, we see that the lower quartile is 16, the upper quartile is 25 and thus the IQR = 25 - 16 = 9.
2. The frequencies of kidney failure are summarized in the following table, by race and city of residence.
| Miami | New York | Los Angeles | Total | |
|---|---|---|---|---|
| Blacks | 56 | 63 | 32 | 151 |
| Whites | 76 | 93 | 98 | 267 |
| Hispanics | 64 | 32 | 50 | 146 |
| Total | 196 | 188 | 180 | 564 |
a) [2]. What is the probability that a patient resides in Los Angeles?
Answer: P(Los Angeles) = 180/564 = .319
b) [2]. What is the probability that a patient is white and lives in New York?
Answer: P(white & New York) = 93/564 = .165
c) [1]. What is the conditional probability that the patient is Hispanic if he or she resides in Miami?
Answer: P(Hispanic | Miami) = 64/196 = .327
d) [1]. Explain whether or not a patient being Hispanic is independent of the patient living in Miami.
Answer: P(Hispanic | Miami) = .327 is unequal to .259 = 146/564 = P(Hispanic). Therefore, being hispanic is not independent of living in Miami because the conditional probability is unequal to the corresponding marginal probability.
Or, compare joint probability, P(Hispanic & Miami ) 64/564 = .113 with the product of the marginals P(Hispanic)P(Miami) = (146/564)(196/564) = .089. Since these are unequal, being Hispanic is not independent of living in Miami.
3. Here are model cummulative distribution functions for the ``Age smokers first tried a cigarette'' (upper, darker curve) and for the ``Age smokers began smoking daily'' (lower, lighter curve), as they appeared in the St. Louis Post Dispatch of February 8, 1998.
Answer: Median age = X.5 = 17 approximately.
b) [2]. What percentage of smokers began smoking daily between the ages of 14 and 16?
Answer: Percentage = P(14 <= X <= 16) = 35 - 15 = 20%, roughly.
c) [2]. Estimate the 80th percentile age at which smokers first tried a cigarette.
Answer: Y.8 = 17, roughly.
4. Faculty salaries at a midwestern university are normally distributed with a mean of $51,500 and standard deviation of $3,000.
a) [3]. Find the probability that one faculty member chosen at random has a salary less than $50,000.
Answer: X = salary of a randomly selected faculty member.
Given that X ~ N(51500, 3000), normal with mean 51,500 and standard deviation 3,000.
P(X < 50000) = P(Z < (50000 - 51500)/3000) = P(Z <= -.5) = .3085 by Table A2* or calculator.
The answer is found on the TI83 with normalcdf(-10^99,50000,51500,3000) = .3085
b) [1]. What is the probability that the average salary of four faculty members chosen at random is less than $50,000?
Answer: Xbar = sample mean for sample of size n = 4.
X ~ N(51500, 3000) implies Xbar ~ N(51500, 3000/sqrt(4)) = N(51500, 1500)
P(Xbar < 50000) = P(Z < (50000 - 51500)/1500) = P(Z <= -1) = .1587
5. Of the vehicles inspected at an emissions check, 10% have emissions problems.
a) [3]. For a random sample of 20 vehicles, what is the probability that at most 3 have emissions problems?
Answer: X = number of emissions problems in 20 vehicles.
Given X ~ B(20, .1).
P(X <= 3) = .8670 by Table A1.
Here np = 20x.1 = 2 is not >= 5, so the normal approximation is not recommended. It gives the approximation P(Z <= (3 + .5 - 2)/sqrt(1.8)) = .8682
b) [1]. For a random sample of 50 vehicles, what is the probability that at least 5 have emissions problems?
Answer: Y = number of emissions problems in 50 vehicles.
Y ~ B(50, .1). Table A1 doesn't include n = 50. We must use the normal approximation or a good calculator.
Normal approximation: P(Y >= 5) = 1 - P(Y <= 4) = 1 - P(Z <= (4 + .5 - 5)/sqrt(4.5)) = 1 - P(Z <= -.236) = 1 - .4068 = .5932
With the TI83, P(Y >= 5) = 1 - P(Y <= 4) = 1 - binomcdf(50,.1,4) = .5688
Notice that the normal approximation isn't too good in this case.
6. Sue is one of five applicants for a job. After completion of all the interviews, a manager ranks the applicants from 1 (most desirable) to 5 (least desirable). Assume all five applicants for the position are equally capable, so Sue has an equal chance of being ranked first, second, third, fourth or fifth. Let X1 be the rank Sue gets from this manager.
a) [2]. What is the mean mu1 and standard deviation sigma1 of X1?
Answer: Values of X1 are 1, 2, 3, 4, 5, each with probability .2
mu1 = (1+2+3+4+5)/5 = 3
sigma1 = sqrt((1-3)2+(2-3)2+(3-3)2+(4-3)2+(5-3)2)/5) = sqrt(2) = 1.414
b) [1]. What is the probability that Sue's rank is greater than 3?
Answer: P(X1 > 3) = .2 + .2 = 2/5 = .4
c) [1]. Seven additional managers also interview and rank the applicants, independent of the first manager and of each other. Let Xi be the rank the ith manager gives to Sue, for i=1,2,...,8. Let Xbar be the average of her ranks from these 8 managers. What is the mean and standard deviation of Xbar?
Answer: mubar = mu1 = 3
sigmabar = sigma1/sqrt(8) = .5
d) [1]. Use the Central Limit Theorem to estimate the probability that this average of Sue's ranks is greater than 3.
Answer: C.L.T. implies that Xbar is approximately distributed as N(3,.5). Therefore, P(Xbar > 3) = 1 - P(Z <= (3-3)/.5) = .5