Question

Suppose 2' is a normally distributed random variable with ft = 10.3 and 0 = 3.8. For the following probability,draw an appropriate diagram, shade the appropriate region and then determine the value:P(9 <2 ≤ 14) = Note: Enter your answer up to 4 decimal places.

Answer 1

GIVEN

The following values are given:

$\begin{gathered} \mu=10.3 \\ \sigma=3.8 \end{gathered}$

SOLUTION

The z-score for the x values 9 and 14 can be calculated using the formula:

$z=\frac{x-\mu}{\sigma}$

For x = 9:

$\begin{gathered} z=\frac{9-10.3}{3.8} \\ z=-0.34 \end{gathered}$

For x = 14:

$\begin{gathered} z=\frac{14-10.3}{3.8} \\ z=0.97 \end{gathered}$

The probability can be calculated as follows:

$P(9\le x\le14)=Pr(-0.34The region that represents the solution is shown below:

Therefore, the probability is given to be:

[tex]P(9\le x\le14)=0.4671$

The probability is 0.4671.