Question

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 85 hours and a standard deviation of 11 hours. Complete parts a through c.

a. What is the probability that a single battery randomly selected from the population will have a life between 80

and 90​ hours?

​P(80≤ overbar x≤90​)= ​(Round to four decimal places as​ needed.)

b. What is the probability that 4 randomly sampled batteries from the population will have a sample mean life of between 80 and 90 hours?

​P(80≤overbar x≤90​)= ​(Round to four decimal places as​ needed.)

c. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 11to9 ​hours, what would be the probability that 4 batteries randomly sampled from the population will have a sample mean life of between 80 and 90 hours?

​P(80≤overbar x≤90​)= ​(Round to four decimal places as​ needed.)

Answer 1

a. 0.3472
(U = 85, sd = 11; n = 1; √n = √1 = 1
P(80 < X < 90) = P(X ≤ 90) - P(X ≤ 80) = 0.6736 - 0.3264 = 0.3472
P(X≤90): z = (X - U)/(sd/√n) = (90-85)/11 = 5/11 = 0.45
P(z = 0.45) = 0.6736
P(X≤80): z = (X - U)/sd = (80-85)/11 = -5/11 = -0.45
P(z = -0.45) = 0.3264)

b. 0.6372
(U = 85, sd = 11; n = 4; √n = √4 = 2; sd/√n = 11/2 = 5.5
P(80 < X < 90) = P(X ≤ 90) - P(X ≤ 80) = 0.8186 - 0.1814 = 0.6372
P(X≤90): z = (X - U)/(sd/√n) = (90-85)/5.5 = 5/5.5 = 0.91
P(z = 0.91) = 0.8186
P(X≤80): z = (X - U)/sd = (80-85)/5.5 = -5/5.5 = -0.91
P(z = -0.91) = 0.1814)

c. 0.7330
(U = 85, sd = 9; n = 4; √n = √4 = 2; sd/√n = 9/2 = 4.5
P(80 < X < 90) = P(X ≤ 90) - P(X ≤ 80) = 0.8665 - 0.1335 = 0.7330
P(X≤90): z = (X - U)/(sd/√n) = (90-85)/4.5 = 5/4.5 = 1.11
P(z = 1.11) = 0.8665
P(X≤80): z = (X - U)/sd = (80-85)/4.5 = -5/4.5 = -1.11
P(z = -1.11) = 0.1335)