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

Answer:

$z = \frac{981.7 -997}{\frac{62}{\sqrt{73}}}= -2.108$

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

$P(z>-2.108) =1- P(z$

Step-by-step explanation:

We know the following info given:

$\mu = 997$ represent the true mean

$\sigma = \sqrt{3844}= 62$ 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:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

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:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{981.7 -997}{\frac{62}{\sqrt{73}}}= -2.108$

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

$P(z>-2.108) =1- P(z$