Question

The standard deviation of the scores on a skill evaluation test is 354354 points with a mean of 14581458 points. If 313313 tests are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 4040 points? Round your answer to four decimal places.

Answer 1

Answer:

$z =\frac{1418-1458}{\frac{354}{\sqrt{313}}}= -2.0$

$z =\frac{1498-1458}{\frac{354}{\sqrt{313}}}= 2.0$

And we can find this probability with this difference and using the normal standard distribution or excel and we got:

$P(-2<z<2) = P(z<2) -P(z<2)= 0.977-0.0228 = 0.9542$

Step-by-step explanation:

For this case we know the following parameters:

$\mu = 1458, \sigma = 354$

We select a sample size of n = 313 and we want to find the following probability:

$P( 1458- 40 <\bar X < 1458 + 40)= P(1418< \bar X < 1498)$

And we can use the z score formula given by:

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

And using this formula we have:

$z =\frac{1418-1458}{\frac{354}{\sqrt{313}}}= -2.0$

$z =\frac{1498-1458}{\frac{354}{\sqrt{313}}}= 2.0$

And we can find this probability with this difference and using the normal standard distribution or excel and we got:

$P(-2<z<2) = P(z<2) -P(z<2)= 0.977-0.0228 = 0.9542$