Answer:
The correct option is D
Step-by-step explanation:
From the question we are told that
The standard deviation is [tex]\sigma = 16[/tex]
The sample size is n = 64
The standard error of mean is mathematically evaluated as
[tex]\sigma _{\= x } = \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]\sigma _{\= x } = \frac{16 }{\sqrt{64} }[/tex]
[tex]\sigma _{\= x } = 2[/tex]
Generally the probability that the sample mean will be within 2 of the population mean is mathematically represented as
[tex]P( \mu - 2 < \= x < \mu + 2) = P(\frac{( \mu - 2 ) - \mu }{\sigma_{\= x }} < \frac{ \= x - \mu }{\sigma_{\= x }} < \frac{( \mu +2 ) - \mu }{\sigma_{\= x }} )[/tex]
Generally [tex]\frac{ \= x - \mu }{\sigma_{\= x }} = Z (The \ standardized \ value \ of \ \= x )[/tex]
So
[tex]P( \mu - 2 < \= x < \mu + 2) = P(\frac{( \mu - 2 ) - \mu }{\sigma_{\= x }} < Z< \frac{( \mu +2 ) - \mu }{\sigma_{\= x }} )[/tex]
[tex]P( \mu - 2 < \= x < \mu + 2) = P(\frac{( -2 }{\sigma_{\= x }} < Z< \frac{ 2 }{\sigma_{\= x }} )[/tex]
substituting values
[tex]P( \mu - 2 < \= x < \mu + 2) = P(\frac{-2 }{2} < Z< \frac{ 2 }{2} )[/tex]
[tex]P( \mu - 2 < \= x < \mu + 2) = P(-1< Z< 1 )[/tex]
=> [tex]P( \mu - 2 < \= x < \mu + 2) = P(Z < 1) - P(Z < -1)[/tex]
From the normal distribution table [tex]P(Z < 1 ) = 0.84134[/tex]
[tex]P(Z < - 1) = 0.15866[/tex]
=> [tex]P( \mu - 2 < \= x < \mu + 2) = 0.84134 - 0.15866[/tex]
=> [tex]P( \mu - 2 < \= x < \mu + 2) = 0.6826[/tex]
