Question

given a normal distribution with μ = 105 and σ = 20, and given you select a sample of n = 16, complete parts (a) through (d)
a. what is the probability that \\( \overline { x } \\) is less than 91?
\\( p ( \overline { x } < 91 ) = 0.0026 \\)
(type an integer or decimal rounded to four decimal places as needed)
b. what is the probability that \\( \overline { x } \\) is between 91 and 93.5?
\\( p ( 91 < \overline { x } < 93.5 ) = 0.0081 \\)
(type an integer or decimal rounded to four decimal places as needed)
c. what is the probability that \\( \overline { x } \\) is above 105.6?
\\( p ( \overline { x } > 105.6 ) = 0.4522 \\)
(type an integer or decimal rounded to four decimal places as needed)
d. there is a 60% chance that \\( \overline { x } \\) is above what value?
\\( \overline { x } = \square \\)
(type an integer or decimal rounded to two decimal places as needed)

Explanation:

Step1: Find the z - score formula for the sample mean

The formula for the z - score of the sample mean \(\bar{X}\) is \(z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\mu = 105\), \(\sigma=20\), and \(n = 16\). So, \(\frac{\sigma}{\sqrt{n}}=\frac{20}{\sqrt{16}}=\frac{20}{4} = 5\).
We know that \(P(\bar{X}>x)=0.6\), then \(P(\bar{X}\leq x)=1 - 0.6=0.4\).

Step2: Look up the z - score in the standard normal distribution table

Looking up the \(z\) - value in the standard normal distribution table (\(Z\sim N(0,1)\)) for which \(P(Z\leq z)=0.4\). The \(z\) - score corresponding to a cumulative probability of \(0.4\) is approximately \(z=- 0.25\) (using the standard normal table or a calculator with a normal - distribution function).

Step3: Solve for \(\bar{X}\)

Using the z - score formula \(z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\), and substituting \(z=-0.25\), \(\mu = 105\), and \(\frac{\sigma}{\sqrt{n}} = 5\).
We get \(-0.25=\frac{\bar{X}-105}{5}\).
Multiply both sides by \(5\): \(-0.25\times5=\bar{X}-105\).
\(-1.25=\bar{X}-105\).
Add \(105\) to both sides: \(\bar{X}=105-1.25\).

Answer:

\(103.75\)