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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.

Suppose 2' is a normally distributed random variable with ft = 10.3 and 0 = 3.8. For-example-1

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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=(x-\mu)/(\sigma)

For x = 9:


\begin{gathered} z=(9-10.3)/(3.8) \\ z=-0.34 \end{gathered}

For x = 14:


\begin{gathered} z=(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:<p>Therefore, the probability is given to be:</p>[tex]P(9\le x\le14)=0.4671

The probability is 0.4671.

Suppose 2' is a normally distributed random variable with ft = 10.3 and 0 = 3.8. For-example-1
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