Question

a simple random sample of size ( n = 64 ) is obtained from a population that is skewed right with ( mu = 88 ) and ( sigma = 32 ).
(a) describe the sampling distribution of ( \bar{x} ).
(b) what is ( p(\bar{x}>95.4) )?
(c) what is ( p(\bar{x}leq78.6) )?
(d) what is ( p(86<\bar{x}<97.8) )?
(a) choose the correct description of the shape of the sampling distribution of ( \bar{x} ).
a. the distribution is approximately normal.
b. the distribution is uniform.
c. the distribution is skewed left.
d. the distribution is skewed right.
e. the shape of the distribution is unknown.
find the mean and standard deviation of the sampling distribution of ( \bar{x} ).
( mu_{\bar{x}} = 88 )
( sigma_{\bar{x}} = 4 )
(type integers or decimals. do not round.)
(b) ( p(\bar{x}>95.4)=0.0322 ) (round to four decimal places as needed.)
(c) ( p(\bar{x}leq78.6)=square ) (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\). We know that \(\mu_{\bar{x}} = 88\), \(\sigma_{\bar{x}}=4\), and \(\bar{x}=78.6\).

$$z=\frac{78.6 - 88}{4}=\frac{-9.4}{4}=-2.35$$

Step2: Find the probability

We want to find \(P(\bar{X}\leq78.6)\), which is equivalent to \(P(Z\leq - 2.35)\) using the standard normal distribution. Looking up the value of \(z =-2.35\) in the standard normal table (or using a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(-2.35,TRUE)), we get the probability.

Answer:

\(P(\bar{X}\leq78.6)=0.0094\)