Statistics Exercises

Exercise 6-2: Standard Normal Distribution

Exercise 6-2

Standard Normal Distribution. In Exercises 10, find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.
Standard Normal Distribution. In Exercises 13 and 16, find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.
54. If a continuous uniform distribution has parameters of m = 0 and s = 1, then the minimum is – 2 3 and the maximum is 2 3
a.    For this distribution, find P (-1 6 x 6 1).
b.    Find P (-1 6 x 6 1) if you incorrectly assume that the distribution is normal instead of uniform.
c.    Compare the results from parts (a) and (b). Does the distribution affect the results very much?

Exercise 6-3
IQ Scores. In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test).
IQ Scores. In Exercises 9–12, find the indicated IQ score. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test).

Exercise 6-5

1.    Population Parameters Three randomly selected households are surveyed as a pilot project for a larger survey to be conducted later. The numbers of people in the households are 2, 3, and 10 (based on Data Set 22 in Appendix B). Consider the values of 2, 3, and 10 to be a population. Assume that samples of size n = 2 are randomly selected without replacement.
a.    Find m and s.
b.    After finding all samples of size n = 2 that can be obtained without replacement, find the population of all values of x by finding the mean of each sample of size n = 2.
c.    Find the mean mx and standard deviation sx for the population of sample means found in
d.    Verify that
mx = m and   s- =   s/n   N  -  n/N-1

Exercise 6-6

1.    M and S Multiple choice test questions are commonly used for standardized tests, including the SAT, ACT, and LSAT. When scoring such questions, it is common to compensate for guessing. If a test consists of 100 multiple choice questions, each with possible answers of a, b, c, d, e, and each question has only one correct answer, find m and s for the number of correct answers provided by someone who makes random guesses. What do m and s measure?
Applying Continuity Correction. In Exercises 5–12, the given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. For example, the probability of “more than 20 defective items” corresponds to the area of the normal curve de- scribed with this answer: “the area to the right of 20.5.”
2.    Probability of more than 8 Senators who are women
3.    Probability of at least 2 traffic tickets this year
4.    Probability of fewer than 5 passengers who do not show up for a flight
5.    Probability that the number of students who are absent is exactly 4
6.    Probability of no more than 15 peas with green pods
7.    Probability that the number of defective computer power supplies is between 12 and 16 inclusive
8.    Probability that the number of job applicants late for interviews is between 5 and 9 inclusive
9.    Probability that exactly 24 felony indictments result in convictions

Exercise 6-3: IQ Scores

Using Normal Approximation. In Exercises 13–16, do the following: (a) Find the indicated binomial probability by using Table A-1 in Appendix A. (b) If np » 5 and nq » 5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if np<5 or nq<5, then state that the normal approximation is not suitable.
With n  =  10 and p  =  0.5, find P (3).
With n  =  12 and p  =  0.8, find P (9).
With n = 8 and p = 0.9, find P (at least 6).
With n = 15 and p = 0.4, find P (fewer than 3).

Exercise 7-3
1.    Confidence Interval Based on the heights of women listed in Data Set 1 in Appendix B, and assuming  that heights of women have a standard deviation of s = 2.5 in., this 95% confidence interval is obtained: 62.42 in. 6 m 6 63.97 in. Assuming  that you are a newspaper reporter, write a statement that correctly interprets that confidence interval and includes all of the relevant information.

Exercise 7-4
Using Correct Distribution. In Exercises 5–12, assume that we want to construct a confidence interval using the given confidence level. Do one of the following, as appropriate: (a) Find the critical value    A>2, (b) find the critical value   A>2,
 
state that neither the normal nor the t distribution applies.

1.    95%; n = 23; s is unknown; population appears to be normally distributed.
2.    99%; n = 25; s is known; population appears to be normally distributed.
3.    99%; n = 6; s is unknown; population appears to be very skewed.
4.    95%; n = 40; s is unknown; population appears to be skewed.
5.    90%; n = 200; s = 15.0; population appears to be skewed.
6.    95%; n = 9; s is unknown; population appears to be very skewed.
7.    99%; n = 12; s is unknown; population appears to be normally distributed.
8.    95%; n = 38; s is unknown; population appears to be skewed.

Exercise  7-5 

Finding Critical Values. In Exercises 5–8, find the critical values x2 and x2 that correspond to the given confidence level and sample size.
5. 95%; n = 9
7. 99%; n = 81
Exercise  7-2
1.    Sampling Suppose the poll results from Exercise 1 were obtained by mailing 100,000 questionnaires and receiving 21,944 responses. Is the result of 43% a good estimate of the population percentage of “yes” responses? Why or why not?

2.    Poll Results in the Media USA Today provided a “snapshot” illustrating poll results from 21,944 subjects. The illustration showed that 43% answered “yes” to this question: “Would you rather have a boring job than no job?” The margin of error was given as ;1 percentage point. What important feature of the poll was omitted?

3.     Find za 2 for a = 0.10-2.

4.    98% confidence; the sample size is 1230, of which 40% are successes.

5.    22. n = 2000, x = 400, 95% confidence.

6.    Margin of error: three percentage points; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 87%.

7.    Postponing Death An interesting and popular hypothesis is that individuals can temporarily postpone their death to survive a major holiday or important event such as a birthday. In a study of this phenomenon, it was found that in the week before and the week after Thanksgiving, there were 12,000 total deaths, and 6062 of them occurred in the week before Thanksgiving (based on data from “Holidays, Birthdays, and Postponement of Cancer Death,” by Young and Hade, Journal of the American Medical Association, Vol. 292, No. 24.)

a.    What is the best point estimate of the proportion of deaths in the week before Thanksgiving to the total deaths in the week before and the week after Thanksgiving?
b.    Construct a 95% confidence interval estimate of the proportion of deaths in the week be- fore Thanksgiving to the total deaths in the week before and the week after Thanksgiving.
c.    Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the Thanksgiving holiday? Why or why not?

8.    Misleading Survey Responses In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.
a.    Find a 99% confidence interval estimate of the proportion of people who say that they voted.
b.    Are the survey results consistent with the actual voter turnout of 61%? Why or why not?

9.    Margin of error: 0.045; confidence level: 95%; and unknown.