Question

A manufacturer of automobile batteries claims that the distribution of the lengths of life of its battery has mean of 54 months and a standard deviation of 6 months. Recently manufacturer has received a rash of complaints from unsatisfied customers whose batteries have died earlier than expected. Suppose a consumer group decides to check the manufacturer’s claim by purchasing a sample of 50 of these batteries and subjecting them to test that determine battery life. Assuming that the manufacturer claim is true,

(a) What is the mean of the sample mean lengths of life of the batteries?

(b) What is the standard deviation of the sample mean lengths of life of the batteries?

(c) What is the variance of the sample mean lengths of life of the batteries?

(d) What is the sampling distribution of the sample mean lengths of life of the batteries?

(e) What is the probability that the of the sample mean lengths of life of the batteries is greater than 60 months?

(f) What is the probability that of the sample mean lengths of life of the batteries is less than 65 months?

(g) What is the probability that of the sample mean lengths of life of the batteries is between 55 to 65 months?

Answer 1

a) = 54

b)

= 6/ = 0.8485

c) = () ^​​​2 = 0.72

d) The sampling distribution of the sample mean is approximately normally distributed with mean equal to 54 and standard deviation of 0.8485.

e) P( > 60)

= P(( - )/() > (60 - )/())

= P(Z > (60 - 54)/0.8485)

= P(Z > 7.07)

= 1 - P(Z < 7.07)

= 1 - 1 = 0

f) P( < 65)

= P(( - )/() < (65 - )/())

= P(Z < (65 - 54)/0.8485)

= P(Z < 12.96)

= 1

G) P(55 < < 65)

= P(55 - )/() < ( - )/() < (65 - )/() )

= P((55 - 54)/0.8485 < Z < (65 - 54)/0.8485

= P(1.18 < Z < 12.96

= P(Z < 12.96) - P(Z < 1.18)

=