Question

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. (a) What are the mean and standard deviation of the sampling distribution? μx = 40 σx = 0.635 (b) What is the approximate probability that x will be within 0.4 of the population mean μ? (Round your answer to four decimal places.) P =

Answer 1

Answer:

a)  $\bar X \sim N(40,\frac{5}{\sqrt{64}}=0.625)$

b) $P(39.6 \leq \bar X \leq 40.4)=0.4778$

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".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=40,\sigma=5)$  

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

(a) What are the mean and standard deviation of the sampling distribution?

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

On this case  $\bar X \sim N(40,\frac{5}{\sqrt{64}}=0.625)$

(b) What is the approximate probability that x will be within 0.4 of the population mean μ? (Round your answer to four decimal places.) P =

So for this case we want this probability:

$P(40-0.4 \leq \bar X \leq 40+0.4)= P(39.6 \leq \bar X \leq 40.4)$

And for this case we can use the z score given by this formula:

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

And using this concept we got this:

$P(\frac{39.6 -40}{\frac{5}{\sqrt{64}}} \leq \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}} \leq \frac{40.4 -40}{\frac{5}{\sqrt{64}}})$

$P(-0.64 \leq Z \leq 0.64) =P(z$