Sect. 5-2, p. 209 Identifying ProbabilityDistributions. In Exercise 7-12, determine whether a probability distribution is given. In those cases where a probability distribution is not described, identify the requirements that are not satisfied. In those cases where a probability distribution is described, find its meanand standard deviation. (see example on p. 206, 207)

#7. Genetic Disorder. Three males with an X-linked genetic disorder have one child each. The random variablex is the number of children among the three who inherit the X-linked genetic disorder.

X P(x)

0 0.4219

1 0.4219

2 0.1406

3 0.0156

Answer: µ = σ =

8. Numbers of Girls. A researcher reports that when groups of four children are randomly selected from a population of couples meeting certain criteria, the probability distribution for the number of girls is as given in the accompanying table.

X P(x)

0 0.502

1 0.365

2 0.098

3 0.011

4 0.001

Answer: µ = σ =

10. Mortality Study. For a group of four men, the probability distribution for the number x who live through the next year is as given in the accompanying table.

X P(x)

0 0.0000

1 0.0001

2 0.0006

3 0.0387

4 0.9606

Answer: µ = σ =

#19a,b (see example on p. 209) Expected Value for a Life Insurance Policy. The CNA Insurance Company charges a 21-year old male a premium of $250 for one year $100,000 life insurance policy. A 21-year old male has a 0.9985 probability of living for a year (based on datafrom the National Center for Health Statistics).

a. From the perspective of a 21-year old male (or his estate), what are the values of the two different outcomes?

Answer:

b. What is the expected value for a 21-year old male who buys the insurance?

Answer:

Sect. 5-3, p. 220 In Exercises 15-20 assume that a procedure yields a binomialdistribution with a trial repeated n times. Use Table A-1 to find the probability of x successes given the probability p of success on a given trial.

#19. n = 14, x = 2, p = 0.30

Answer:

20. n = 15, x= 12, p = 0.90

Answer:

#28 [Hint: P(at least 1) = 1 – P(0)] Using Computer Results. In Exercises 25-28, refer to the Minitab display below. The probabilities obtained by entering the values of n = 6 and p = 0.167. In a clinical test of the drug Lipitor, 16.7% of the subjects treated with 10 mg of atorvastatin experienced headaches (based on data from Parke-Davis). In each case, assume that 6 subjects are randomly selected and treated with 10 mg of atorvastatin, then find the indicated probability.

Minitab

Binomial with n = 6 and p = 0.167000

X P(X = x)

0.00 0.3341

1.00 0.4019

2.00 0.2014

3.00 0.0538

4.00 0.0081

5.00 0.0006

6.00 0.0000

Find the probability that at least one subject experiences headaches. Is it unusual to have at least one of six subjects experience headaches?

Answer:

Sect. 5-4, p. 227 Finding u, σ, and Unusual Values. In Exercises 5-8, 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 rangerule of thumb to find the minimum usual value µ – 2σ and the maximum usual value µ + 2

σ.

#7. n = 1492, p = 1/4

10. (For these problems, see examples on p. 225, 226). Guessing Answers. Several economicsstudents are unprepared

for a multiple-choice quiz with 25 questions, and all of their answers are guesses. Each question has five possible answers, and only one of them is correct.

a. Find the mean and standard deviation for the number of correct answers for such students.

Answer:

b. Would it be unusual for a student to pass by guessing and getting at least 15 correct answers? Why or why not?

Answer:

Sect. 6-3, p. 266. IQ Scores. In Exercises 5-12, assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). (Hint: Draw a graph in each case.)

#5. Find the probability that a randomly selected adult has an IQ that is less than 130.

7. Find the probability that a randomly selected adult has an IQ between 90 and 110 (referred to as the normal range).

9. Find P10 which is the IQ score separating the bottom 10% from the top 90%.

In Exercises 13-16, use this information (based on data from the National Health Survey):

Men’s heights are normally distributed with mean 69.0 in. and standard deviation 2.8 in.

Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.

13. Beanstalk Club Height Requirement. The Beanstalk Club, a social organization for tall people, has a requirement that

women must be at least 70 in. (or 5 ft. 10 in.) tall. What percentage of women meet that requirement?

17. Birth Weights. Birth weights in the US are normally distributed with a mean of 3420 g and a standard deviation of 495 g.

If a hospital plans to set up special observation conditions for the lightest 2% of babies, what weight is used for the cutoff separating the lightest 2% from the others?

18. (For these problems, see p. 260-265) Birth Weights. Birth weights in Norway are normally distributed with a mean of

3570 g and a standard deviation of 500 g. Repeat Exercise 17 for babies born in Norway. Is the result very different from the result found in Exercise 17?

The solution provides step by step method for the calculation of calculation of probability and descriptive statistics in Minitab . Formula for the calculation and Interpretations of the results are also included.