Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 9 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 70 and 80 hours? P(70 < or = x overbar < or = 80) = 0.4246 (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 70 and 80 hours? P(70 < or = x overbar < or = 80) = (Round to four decimal places as needed.
Solution:
Part a
We are given
µ = 75
σ =9
We have to find P(70 < X < 80)
P(70 < X < 80) = P(X<80) – P(X<70)
Find P(X<80)
Z = (X - µ)/σ
Z = (80 – 75)/9
Z = 0.555555556
P(Z< 0.555555556) = P(X<80) = 0.710742639
(by using excel/z-table)
Now find P(X<70)
Z = (X - µ)/σ
Z = (70 – 75)/9
Z = -0.555555556
P(Z< -0.555555556) = P(X<70) = 0.289257361
(by using excel/z-table)
P(70 < X < 80) = P(X<80) – P(X<70)
P(70 < X < 80) = 0.710742639 - 0.289257361
P(70 < X < 80) = 0.421485278
Required probability = 0.4215
Part b
We are given
µ = 75
σ =9
n = 4
We have to find P(70 < Xbar < 80)
P(70 < Xbar < 80) = P(Xbar<80) – P(Xbar<70)
Find P(Xbar<80)
Z = (X - µ)/[σ/sqrt(n)]
Z = (80 – 75)/[9/sqrt(4)]
Z = 1.111111
P(Z< 1.111111) = P(Xbar<80) = 0.86674
(by using excel/z-table)
Now find P(Xbar<70)
Z = (X - µ)/[σ/sqrt(n)]
Z = (70 – 75)/[9/sqrt(4)]
Z = -1.111111
P(Z< -1.111111) = P(Xbar<70) = 0.13326
(by using excel/z-table)
P(70 < Xbar < 80) = P(Xbar<80) – P(Xbar<70)
P(70 < Xbar < 80) = 0.86674 - 0.13326
P(70 < Xbar < 80) = 0.73348
Required probability = 0.7335
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