Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 6262 minutes with a mean life of 606606 minutes. If the claim is true, in a sample of 9999 batteries, what is the probability that the mean battery life would be greater than 619619 minutes

Answer 1

`\bar X \sim N(\mu. (\sigma)/(√(n)))`

And we want to find the following probability:

`P(\bar X >619)`

And we can use the z score formula given by:

`z=(\bar x -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z=(619-606)/((62)/(√(99)))= 2.086`

And we can use the normal standard distirbution and the complement rule to find the probability:

`P(z>2.086)=1 -P(z<2.086)= 1-0.982= 0.018`

Explanation:

For this problem we have the following parameters given:

`\mu = 606` represent the mean

`\sigma = 62` represent the true deviation

`n= 99` represent the sample size

For this case since the sample size is >30 we can use the central limit theorem and we can u se the following distribution for the sample mean

`\bar X \sim N(\mu. (\sigma)/(√(n)))`

And we want to find the following probability:

`P(\bar X >619)`

And we can use the z score formula given by:

`z=(\bar x -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z=(619-606)/((62)/(√(99)))= 2.086`

And we can use the normal standard distirbution and the complement rule to find the probability:

`P(z>2.086)=1 -P(z<2.086)= 1-0.982= 0.018`