Question

1. Suppose the lifespan of a particular bran of calculator battery, X, is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. Please answer A-C

(a) Let x̄ be the mean lifespan of a sample of n batteries. Does x̄ follow a normal distribution? Explain

(b) What is the probability that a single randomly selected battery from the population will have a lifespan between 70 and 80 hours?

(c) What is the probability that 16 randomly selected batteries from the population will have a mean lifespan between 70 and 80 hours?

Answer 1

Part a

We know that the sampling distribution of the sample mean follows an approximate normal distribution with the mean equal to the population mean and standard deviation equal to the standard error of the mean. So, for the given scenario, Xbar follow a normal distribution.

Part b

We are given

µ = 75

σ = 10

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)/10

Z = 0.5

P(Z< 0.5) = P(X<80) = 0.691462

(by using z-table or excel)

Now find P(X<70)

Z = (70 – 75)/10

Z = -0.5

P(Z<-0.5) = P(X<70) = 0.308538

(by using z-table or excel)

P(70<X<80) = P(X<80) – P(X<70)

P(70<X<80) = 0.691462 - 0.308538

P(70<X<80) = 0.382924

Required probability = 0.382924

Part c

We are given

µ = 75

σ = 10

n = 16

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)/[10/sqrt(16)]

Z = 2

P(Z< 2) = P(Xbar<80) = 0.97725

(by using z-table or excel)

Now find P(Xbar<70)

Z = (70 – 75)/[10/sqrt(16)]

Z = -2

P(Z<-2) = P(Xbar<70) = 0.02275

(by using z-table or excel)

P(70<Xbar<80) = P(Xbar<80) – P(Xbar<70)

P(70<Xbar<80) = 0.97725 - 0.02275

P(70<Xbar<80) = 0.9545

Required probability = 0.9545