Question

LongLast Inc. produces car batteries. The mean life of these batteries is 60 months. The distribution of the battery life closely follows the normal probability distribution with a standard deviation of eight months. As a part of its testing program, LongLast tests a sample of 25 batteries. What proportion of the samples will have a mean useful life between 58 and 62 months?

Answer 1

50% of the samples will be in that range (58, 62)

Explanation:

We have that the mean (m) is equal to 60 and the standard deviation is equal to 8. Also that the sample size n = 25

To calculate the value of z, it is the following formula:

z = (x - m) / sd / (n ^ (1/2))

for x = 58:

we replace

z = (58 - 60) / 8 / (25 ^ (1/2))

z = - 0.05

for x = 62:

and when z has this value, the probability is 0.4801

we replace

z = (62 - 60) / 8 / (25 ^ (1/2))

z = 0.05

and when z has this value, the probability is 0.5199

If we average these values, (0.4801 + 0.5199) / 2 = 0.5

Which means that 50% of the samples will be in that range