Question

a simple random sample of size ( n = 42 ) is obtained from a population with ( mu = 64 ) and ( sigma = 17 ).
(a) what must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? assuming that this condition is true, describe the sampling distribution of ( overline{x} ).
(b) assuming the normal model can be used, determine ( p(overline{x} < 68.2) ).
(c) assuming the normal model can be used, determine ( p(overline{x} geq 65.7) ).
assuming the normal model can be used, describe the sampling distribution ( overline{x} ). choose the correct answer below.
a. approximately normal, with ( mu_{overline{x}} = 64 ) and ( sigma_{overline{x}}=\frac{17}{sqrt{42}} ).
b. approximately normal, with ( mu_{overline{x}} = 64 ) and ( sigma_{overline{x}} = 17 ).
c. approximately normal, with ( mu_{overline{x}} = 64 ) and ( sigma_{overline{x}}=\frac{42}{sqrt{17}} ).
(b) ( p(overline{x} < 68.2)=) (round to four decimal places as needed).

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 17$ and $n = 42$, then $\sigma_{\bar{x}}=\frac{17}{\sqrt{42}}\approx2.615$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. For $\bar{x}=68.2$, $\mu_{\bar{x}} = 64$, and $\sigma_{\bar{x}}\approx2.615$, we have $z=\frac{68.2 - 64}{2.615}=\frac{4.2}{2.615}\approx1.606$.

Step3: Find the probability

Using the standard normal distribution table (or a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(1.606,TRUE)), we find $P(Z < 1.606)$.

Answer:

$P(\bar{x}<68.2)\approx0.9463$