3 terms

Given that z is a standard normal random variable, compute the following probabilities.

a) P(-1.98 <= z <=.49)

b) P(.52 <= z <= 1.22)

c) P (-1.75 <= z <= -1.04)

a) P(-1.98 <= z <=.49)

b) P(.52 <= z <= 1.22)

c) P (-1.75 <= z <= -1.04)

a. P (Z < -1.98) - P (Z < .49) = .6879 - .0239 = .664

b. P (Z < .52) - P (Z < 1.22) = .8869 - .6985 = .1884

c. P (Z < -1.75) P (Z < -1.04) = .1492 - .0401 = .1091

b. P (Z < .52) - P (Z < 1.22) = .8869 - .6985 = .1884

c. P (Z < -1.75) P (Z < -1.04) = .1492 - .0401 = .1091

The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions.

a) What is the probability of completing the exam in one hour or less?

b) What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes?

c) Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time?

a) What is the probability of completing the exam in one hour or less?

b) What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes?

c) Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time?

a) P (z < 60) = P (z < 60-80/10) = P (z <-2) = 0.0228 or 2.28%

b) P (60 < z < 75) = P (60 - 80/10 < z< 75-80/10) = P (-1.67 < z < -0.42) = 0.3375 - 0.0475 = 0.2897 or 28.97%

c) P (z > 90) = P (z > 90-80/10) = P (z > 0.83) = 1 - P (z < 0.83) = 1 - 0.7967 = 0.2033 or 20.33%

60 * 0.2033 = ~ 12.198 students

b) P (60 < z < 75) = P (60 - 80/10 < z< 75-80/10) = P (-1.67 < z < -0.42) = 0.3375 - 0.0475 = 0.2897 or 28.97%

c) P (z > 90) = P (z > 90-80/10) = P (z > 0.83) = 1 - P (z < 0.83) = 1 - 0.7967 = 0.2033 or 20.33%

60 * 0.2033 = ~ 12.198 students

According to Salary Wizard, the average base salary for a brand manager in Houston, Texas, is $88,592 and the average base salary for a brand manager in Los Angeles, California, is $97,417 (Salary Wizard website, February 27, 2008). Assume that salaries are normally distributed, the standard deviation for brand managers in Houston is $19,900, and the standard deviation for brand managers in Los Angeles is $21,800.

a) What is the probability that a brand manager in Houston has a base salary in excess of $100,000?

b) What is the probability that a brand manager in Los Angeles has a base salary in excess of $100,000?

c) What is the probability that a brand manager in Los Angeles has a base salary of less than $75,000?

d) How much would a brand manager in Los Angeles have to make in order to have a higher salary than 99% of the brand managers in Houston?

a) What is the probability that a brand manager in Houston has a base salary in excess of $100,000?

b) What is the probability that a brand manager in Los Angeles has a base salary in excess of $100,000?

c) What is the probability that a brand manager in Los Angeles has a base salary of less than $75,000?

d) How much would a brand manager in Los Angeles have to make in order to have a higher salary than 99% of the brand managers in Houston?

a) P (z > 100,000) = P (100,000- 88,592/19,900) = P (z = > 0.57) = .2843 or 28.4%

b) P (z > 100,000) = P (100,000 - 97,417/21,800) = P (z = > 0.11) = 0.4522 or 45.2%

c) P (z < 75,000) = P (75,000 - 97,417/21,800) = P (z = < -1.028) = .1539 or 15.4%

d)

b) P (z > 100,000) = P (100,000 - 97,417/21,800) = P (z = > 0.11) = 0.4522 or 45.2%

c) P (z < 75,000) = P (75,000 - 97,417/21,800) = P (z = < -1.028) = .1539 or 15.4%

d)