Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 55 minutes with a mean life of 1141 minutes. If the claim is true, in a sample of 55 batteries, what is the probability that the mean battery life would be greater than 1132.1 minutes? Round your answer to four decimal places.

Answer 1

Answer:

$P(\bar X >1132.1)=P(Z>\frac{1132.1-1141}{\frac{55}{\sqrt{55}}}=-1.2)$

And using the complement rule and a calculator, excel or the normal standard table we have that:

$P(Z>-1.2) =1-P(Z<-1.2)=1-0.115=0.885$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(1141,55)$  

Where $\mu=1141$ and $\sigma=55$

For this case the distribution for the sample mean $\bar X$ is given by:

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

We can find the individual probabilities like this:

$P(\bar X >1132.1)=P(Z>\frac{1132.1-1141}{\frac{55}{\sqrt{55}}}=-1.2)$

And using the complement rule and a calculator, excel or the normal standard table we have that:

$P(Z>-1.2) =1-P(Z<-1.2)=1-0.1151=0.8849$