Question

(4) The time required to complete a final exam in a particular college course is normally distributed with a mean of 75 minutes and a standard deviation of 15 minutes. Answer the following questions: (a) (for a randomly selected student) What is the probability of completing the exam in 1 hour or less? (4) (b) What is the probability a randomly selected student will complete the exam in more than 60 minutes but less than 75 minutes? (4b) (c) Assume that the class has 45 students and that the exam period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time? (round it to a whole number.) Hint: find the probability as a percentage, and then multiply the total number of students by the percentage. (40) (5) Assume that men's weights are normally distributed with 4 = 172 lb and o-29 18. (a) If one man is randomly selected, find the probability that his weight is greater than 180 lb. (5a) (b) If 60 men are randomly selected, find the probability that their mean weight is greater than 180 lb. (56)

Answer 1

4) Population mean, µ = 75

Population standard deviation, σ = 15

a)

Probability of completing exam in one hour(60 minutes) or less =

= P(X <= 60) =

= P( (X-µ)/σ <= (60-75)/15 )

= P(z < -1)

Using excel function:

= NORM.S.DIST(-1, 1)

= 0.1587

b)

P(60 < X < 75) =

= P( (60-75)/15 < (X-µ)/σ < (75-75)/15 )

= P(-1 < z < 0)

= P(z < 0) - P(z < -1)

Using excel function:

= NORM.S.DIST(0, 1) - NORM.S.DIST(-1, 1)

= 0.3413

c)

P(X > 90) =

= P( (X-µ)/σ > (90-75)/15)

= P(z > 1)

= 1 - P(z < 1)

Using excel function:

= 1 - NORM.S.DIST(1, 1)

= 0.1587 = 15.87%

Number of student that were not able to complete the exam on time = 45*15.87%

= 7.14 = 7 students