Question

A population has a standard deviation of 16. If a sample of size 64 is selected from this population, what is the probability that the sample mean will be within 2 of the population mean?
a. Since the mean is not given, there is no answer to this question.

b. -0.6826

c. 0.3413

d. 0.6826

e. -0.3413

Answer 1

Answer:

The correct option is  D

Step-by-step explanation:

From the question we are told that

    The standard deviation is  $\sigma = 16$

     The sample size is  n =  64

The standard error of mean is mathematically evaluated as

        $\sigma _{\= x } = \frac{\sigma }{\sqrt{n} }$

substituting values

        $\sigma _{\= x } = \frac{16 }{\sqrt{64} }$

        $\sigma _{\= x } = 2$

Generally the probability that the sample mean will be within 2 of the population mean is mathematically represented as

              $P( \mu - 2 < \= x < \mu + 2) = P(\frac{( \mu - 2 ) - \mu }{\sigma_{\= x }} < \frac{ \= x - \mu }{\sigma_{\= x }} < \frac{( \mu +2 ) - \mu }{\sigma_{\= x }} )$

Generally  $\frac{ \= x - \mu }{\sigma_{\= x }} = Z (The \ standardized \ value \ of \ \= x )$

So

         $P( \mu - 2 < \= x < \mu + 2) = P(\frac{( \mu - 2 ) - \mu }{\sigma_{\= x }} < Z< \frac{( \mu +2 ) - \mu }{\sigma_{\= x }} )$

         $P( \mu - 2 < \= x < \mu + 2) = P(\frac{( -2 }{\sigma_{\= x }} < Z< \frac{ 2 }{\sigma_{\= x }} )$

substituting values

        $P( \mu - 2 < \= x < \mu + 2) = P(\frac{-2 }{2} < Z< \frac{ 2 }{2} )$

        $P( \mu - 2 < \= x < \mu + 2) = P(-1< Z< 1 )$

=>     $P( \mu - 2 < \= x < \mu + 2) = P(Z < 1) - P(Z < -1)$

From the normal distribution table $P(Z < 1 ) = 0.84134$

                                                          $P(Z < - 1) = 0.15866$

=>  $P( \mu - 2 < \= x < \mu + 2) = 0.84134 - 0.15866$

=>   $P( \mu - 2 < \= x < \mu + 2) = 0.6826$