Suppose you take a random sample from a normally distributed population with mean $\mu$= 175 and compute the sample mean $$\bar { x }$$ . If $$\sigma _ { \bar { x } } ^ { 2 }$$ = 9, find P( $$\bar { x }$$ 172).
A)0300
B)0100
C)1732
D)0577
E)8413

Question

Suppose you take a random sample from a normally distributed population with mean $\mu$= 175 and compute the sample mean $$\bar { x }$$ . If $$\sigma _ { \bar { x } } ^ { 2 }$$ = 9, find P( $$\bar { x }$$ > 172). 
A)0300
B)0100
C)1732
D)0577
E)8413

Quizplus · Accepted Answer

Given $\sigma_{\bar{x}}^2 = 9$, the standard deviation of the sample mean, $\sigma_{\bar{x}}$, is $\sqrt{9} = 3$. To find $P(\bar{x} > 172)$, we use the Z-score formula: $Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$. Substituting $\bar{x} = 172$, $\mu = 175$, and $\sigma_{\bar{x}} = 3$, we get $Z = \frac{172 - 175}{3} = -1$. Looking up the Z-score of -1 in the standard normal distribution table gives the probability of being less than 172, which is 0.1587. Therefore, $P(\bar{x} > 172) = 1 - 0.1587 = 0.8413$.

Suppose You Take a Random Sample from a Normally Distributed μ\muμ

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