Question

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that a single battery randomly selected from the population will have a life?
a) no more than 72 hours
b) at least 77 hours
*what is the probability that 16 randomly sampled batteries from the population will have a sample mean life between 70 and 80 hours?

Answer 1

Answer:

a) $P(X$

And we can find this probability using the normal standard table or excel:

$P(z$

b) $P(X>77)=P(\frac{X-\mu}{\sigma}>\frac{77-\mu}{\sigma})=P(Z>\frac{77-75}{10})=P(z>0.2)$

And we can find this probability using the complement rule and the normal standard table or excel:

$P(z>0.2)=1-P(Z$

c) $z = \frac{70-75}{\frac{10}{\sqrt{16}}}= -2$

$z = \frac{80-75}{\frac{10}{\sqrt{16}}}= 2$

And we can find this probability with this difference:

$P(-2$

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 life of a particular brand of a population, and for this case we know the distribution for X is given by:

$X \sim N(75,10)$  

Where $\mu=75$ and $\sigma=10$

Part a

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table or excel:

$P(z$

Part b

$P(X>77)=P(\frac{X-\mu}{\sigma}>\frac{77-\mu}{\sigma})=P(Z>\frac{77-75}{10})=P(z>0.2)$

And we can find this probability using the complement rule and the normal standard table or excel:

$P(z>0.2)=1-P(Z$

Part c

For this case we select a sample size of n =16. and we want this probability:

$P(70 < \bar X$

We know that the distribution for the sample mean $\bar X$ is given by:

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

And we can find the z scores for the limits like this:

$z = \frac{70-75}{\frac{10}{\sqrt{16}}}= -2$

$z = \frac{80-75}{\frac{10}{\sqrt{16}}}= 2$

And we can find this probability with this difference:

$P(-2$