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)

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