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 ( overline{x} ).
(b) what is ( p(overline{x}>95.4) )?
(c) what is ( p(overline{x}leq78.6) )?
(d) what is ( p(86<overline{x}<97.8) )?
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 ( overline{x} ).
( mu_{overline{x}}= )
( sigma_{overline{x}}= )
(type integers or decimals. do not round.)
(b) ( p(overline{x}>95.4)= )
(c) ( p(overline{x}leq78.6)= )
(d) ( p(86<overline{x}<97.8)= )
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - scores

For \(x_1 = 86\), the z - score \(z_1=\frac{x_1-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{86 - 88}{4}=\frac{- 2}{4}=-0.5\)
For \(x_2 = 97.8\), the z - score \(z_2=\frac{x_2-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{97.8 - 88}{4}=\frac{9.8}{4}=2.45\)

Step2: Use the standard normal distribution table

\(P(86<\bar{x}<97.8)=P(-0.5 < Z < 2.45)\)
Since \(P(-0.5 < Z < 2.45)=P(Z < 2.45)-P(Z < - 0.5)\)
From the standard normal table, \(P(Z < 2.45)=0.9929\) and \(P(Z < -0.5)=0.3085\)

Step3: Calculate the probability

\(P(-0.5 < Z < 2.45)=0.9929-0.3085 = 0.6844\)

Answer:

\(0.6844\)