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How did we get 85.89  and 0.1235 using a calculator to get the correct answer?  Thanks!

How did we get 85.89  and 0.1235 using a calculator to get the correct answer?  Thanks!

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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How did we get 85.89  and 0.1235 using a calculator to get the correct answer?  Thanks!

The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and standard deviation 7. Volunteers in the top 20 percent of hours worked will receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 ? (A) 0.077 (B) 0.123 (C) 0.134 (D) 0.618 (E) 0.923
Transcribed Image Text:The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and standard deviation 7. Volunteers in the top 20 percent of hours worked will receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 ? (A) 0.077 (B) 0.123 (C) 0.134 (D) 0.618 (E) 0.923
(C) Correct. If X represents the number of hours worked, then the value of X for which 20 percent of the hours worked are greater than X in a normal distribution with mean 80 and standard deviation 7 can be found using technology to be approximately 85.89. Then the probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 is given by P(X > 85.89 | X < 90) = P(85.89 < X < 90) P(X < 90) Technology can be used to find that P(85.89 < X <90) ≈ 0.1235 and that P(X<90) ≈ 0.924 in a normal distribution with mean 80 and standard deviation 7, so P(X > 85.89 | X < 90) = P(85.89 < X < 90) P(X <90) 0.1235 0.9234 ≈ 0.134.
Transcribed Image Text:(C) Correct. If X represents the number of hours worked, then the value of X for which 20 percent of the hours worked are greater than X in a normal distribution with mean 80 and standard deviation 7 can be found using technology to be approximately 85.89. Then the probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 is given by P(X > 85.89 | X < 90) = P(85.89 < X < 90) P(X < 90) Technology can be used to find that P(85.89 < X <90) ≈ 0.1235 and that P(X<90) ≈ 0.924 in a normal distribution with mean 80 and standard deviation 7, so P(X > 85.89 | X < 90) = P(85.89 < X < 90) P(X <90) 0.1235 0.9234 ≈ 0.134.
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