Question

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 38 hours and a standard deviation of 5.8 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 9 batteries.a. What can you say about the shape of the distribution of the sample mean?
b. What is the standard error of the distribution of the sample mean? (Round your anser to 4 decimals places.)
c. What proportion of the samples will have a mean useful life of more than 39.5 hours? (Round z value to 2 decimal places and final answer to 4 decimal places)
d. What proportion of the sample will have a mean useful life greater than 37.5? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Answer:

a) For this case we select a sample size of n=9. And we know that the distribution of X is normal so then the distribution for the sample mean is given by:

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

b) $SE = \frac{\sigma}{\sqrt{n}} =\frac{5.8}{\sqrt{9}} =1.9333$

c) $P\bar X >39.5)$

And we can use the z score given by:

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

And if we find the z score for 39.5 we got:

$z = \frac{39.5-38}{\frac{5.8}{\sqrt{9}}}= 0.78$

And using the complement rule we got:

$P(z >0.78) =1-P(Z<0.78) = 1-0.7823= 0.2177$

d) $P\bar X >37.5)$

And we can use the z score given by:

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

And if we find the z score for 37.5 we got:

$z = \frac{37.5-38}{\frac{5.8}{\sqrt{9}}}= -0.26$

And using the complement rule we got:

$P(z >-0.26) =1-P(Z<-0.26) = 1-0.3974= 0.6026$

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

$X \sim N(38,5.8)$  

Where $\mu=38$ and $\sigma=5.8$

Part a

For this case we select a sample size of n=9. And we know that the distribution of X is normal so then the distribution for the sample mean is given by:

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

Part b

The standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}} =\frac{5.8}{\sqrt{9}} =1.9333$

Part c

We want this probability:

$P\bar X >39.5)$

And we can use the z score given by:

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

And if we find the z score for 39.5 we got:

$z = \frac{39.5-38}{\frac{5.8}{\sqrt{9}}}= 0.78$

And using the complement rule we got:

$P(z >0.78) =1-P(Z<0.78) = 1-0.7823= 0.2177$

Part d

We want this probability:

$P\bar X >37.5)$

And we can use the z score given by:

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

And if we find the z score for 37.5 we got:

$z = \frac{37.5-38}{\frac{5.8}{\sqrt{9}}}= -0.26$

And using the complement rule we got:

$P(z >-0.26) =1-P(Z<-0.26) = 1-0.3974= 0.6026$