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 ) = \square \\)
(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 - scores

For $X_1 = 91$, the z - score $z_1=\frac{\bar{X}_1-\mu}{\sigma_{\bar{X}}}=\frac{91 - 105}{5}=\frac{-14}{5}=- 2.8$.
For $X_2 = 93.5$, the z - score $z_2=\frac{\bar{X}_2-\mu}{\sigma_{\bar{X}}}=\frac{93.5 - 105}{5}=\frac{-11.5}{5}=-2.3$.

Step3: Find the probabilities

Using the standard normal distribution table, $P(Z\lt - 2.8)=0.0026$ and $P(Z\lt - 2.3)=0.0107$.
Then $P(91\lt\bar{X}\lt93.5)=P(-2.8\lt Z\lt - 2.3)=P(Z\lt - 2.3)-P(Z\lt - 2.8)$.
Substitute the values: $P(-2.8\lt Z\lt - 2.3)=0.0107 - 0.0026=0.0081$.

Answer:

$P(91\lt\bar{X}\lt93.5)=0.0081$