Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 38443844with a mean life of 997997minutes. If the claim is true, in a sample of 7373batteries, what is the probability that the mean battery life would be greater than 981.7981.7minutes

Answer 1

`z = (981.7 -997)/((62)/(√(73)))= -2.108`

And we can use the normal standard distribution table or excel and with the complement ruler we got:

`P(z>-2.108) =1- P(z<-2.108) =1-0.0175= 0.9825`

Explanation:

We know the following info given:

`\mu = 997` represent the true mean

`\sigma = √(3844)= 62` represent the population deviation

`n = 73` represent the sample size selected

Since the sample size is large enough (n>30) we can use the central limit theorem and the sample mean would have the following distribution:

`\bar X \sim N (\mu, (\sigma)/(√(n)))`

And for this case we want to find this probability:

`P(\bar X >981.7)`

And we can use the z score formula given by:

`z= (\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z = (981.7 -997)/((62)/(√(73)))= -2.108`

And we can use the normal standard distribution table or excel and with the complement ruler we got:

`P(z>-2.108) =1- P(z<-2.108) =1-0.0175= 0.9825`