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given a normal distribution with \\( \\mu = 105 \\) and \\( \\sigma = 2…

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 ) = \square \\)

(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error of the mean

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

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{\bar{X}-\mu}{\sigma_{\bar{X}}}$.
Here, $\bar{X}=91$, $\mu = 105$, and $\sigma_{\bar{X}} = 5$.
So, $z=\frac{91 - 105}{5}=\frac{-14}{5}=-2.8$.

Step3: Find the probability

We want to find $P(\bar{X}<91)$, which is equivalent to $P(Z < - 2.8)$ using the standard normal distribution.
Looking up the value in the standard normal table (or using a calculator with a normal - distribution function), $P(Z < - 2.8)=0.0026$.

Answer:

$0.0026$