Question

given a normal distribution with \\( \mu = 105 \\) and \\( \sigma = 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 ) = \\)
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The standard error $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 20$ and $n = 16$, then $\sigma_{\bar{X}}=\frac{20}{\sqrt{16}}=\frac{20}{4} = 5$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{X}-\mu}{\sigma_{\bar{X}}}$. For $\bar{X}=105.6$, $\mu = 105$, and $\sigma_{\bar{X}} = 5$, we have $z=\frac{105.6 - 105}{5}=\frac{0.6}{5}=0.12$.

Step3: Find the probability

We want to find $P(\bar{X}>105.6)$, which is equivalent to $P(Z > 0.12)$. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq0.12)=0.5478$, then $P(Z > 0.12)=1 - 0.5478=0.4522$.

Answer:

$0.4522$