Question

The variance of the scores on a skill evaluation test is 143,641 with a mean of 1517 points. If 343 tests are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 36 points

Answer 1

Answer:

The probability that the mean of the sample would differ from the population mean by less than 36 points=0.9216

Step-by-step explanation:

We are given that

The variance of the scores on a skill evaluation test=143,641

Mean=1517 points

n=343

We have to find the probability that the mean of the sample would differ from the population mean by less than 36 points.

Standard deviation,$\sigma=\sqrt{143641}$

$P(|x-\mu|<36)=P(|\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}|<\frac{36}{\frac{\sqrt{143641}}{\sqrt{343}}})$

$=P(|Z|<\frac{36}{\sqrt{\frac{143641}{343}}})$

$=P(|Z|<1.76)$

$=0.9216$

Hence,  the probability that the mean of the sample would differ from the population mean by less than 36 points=0.9216