Question

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 37 hours. hours and a standard deviation of 5.9 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 9 batteries.

  

a.	What can you say about the shape of the distribution of the sample mean?

   

Sample mean

b.	What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

   

Standard error

c.	What proportion of the samples will have a mean useful life of more than 39 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.)

   

Probability

d.	What proportion of the sample will have a mean useful life greater than 36 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.)

   

Probability

e.	What proportion of the sample will have a mean useful life between 36 and 39 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.)

   

Probability

Answer 1

Solution:

We are given that the mean life of batteries follows the normal probability distribution.

Population Mean = µ = 37

Population SD = σ = 5.9

Sample size = n = 9

Part a

The shape of the sampling distribution of the sample mean follows approximate normal curve. The mean and standard deviation for the sampling distribution of the sample mean is given as below:

Mean of sampling distribution = µ_x̄ = µ = 37

SD of sampling distribution = σ_x̄ = σ/sqrt(n) = 5.9/sqrt(9) = 5.9/3 = 1.966667

Part b

Standard error = σ/sqrt(n) = 5.9/sqrt(9) = 5.9/3 = 1.966667

Standard error = 1.9667

Part c

Here, we have to find P(Xbar>39)

P(Xbar>39) = 1 – P(Xbar<39)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (39 – 37)/[5.9/sqrt(9)]

Z = 2/1.966667

Z = 1.016949

Z = 1.02

P(Z<1.02) = 0.846136

(by using z-table)

P(Xbar<39) = 0.846136

P(Xbar>39) = 1 – P(Xbar<39)

P(Xbar>39) = 1 – 0.846136

P(Xbar>39) = 0.153864

Required probability = 0.1539

Part d

Here, we have to find P(Xbar>36)

P(Xbar>36) = 1 – P(Xbar<36)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (36 – 37)/[5.9/sqrt(9)]

Z = -1/1.966667

Z = -0.50847

Z = -0.51

P(Z<-0.51) = 0.305026

(by using z-table)

P(Xbar<36) = 0.305026

P(Xbar>36) = 1 – P(Xbar<36)

P(Xbar>36) = 1 – 0.305026

P(Xbar>39) = 0.694974

Required probability = 0.6950

Part e

Here, we have to find P(36<Xbar<39)

P(36<Xbar<39) = P(Xbar<39) – P(Xbar<36)

First find P(Xbar<39)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (39 – 37)/[5.9/sqrt(9)]

Z = 2/1.966667

Z = 1.016949

Z = 1.02

P(Z<1.02) = 0.846136

(by using z-table)

P(Xbar<39) = 0.846136

Now, find P(Xbar<36)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (36 – 37)/[5.9/sqrt(9)]

Z = -1/1.966667

Z = -0.50847

Z = -0.51

P(Z<-0.51) = 0.305026

(by using z-table)

P(Xbar<36) = 0.305026

P(36<Xbar<39) = P(Xbar<39) – P(Xbar<36)

P(36<Xbar<39) = 0.846136 - 0.305026

P(36<Xbar<39) = 0.54111

Required probability = 0.5411