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Given the normal distribution with mean = 47 and standard deviation = 4, what is the probability that: 1. X<38 or X>45 2. 8% of the values are less than what X value. 3. Between what two X values (symmetrically distributed around the mean) are 85% of the values?

Respuesta :

SOLUTION

1. Now, we need to get the z-score for

[tex]\begin{gathered} x<38\text{ and } \\ x>45 \end{gathered}[/tex]

For x < 38, we have

[tex]Z_{38}=\frac{38-47}{4}=-2.25[/tex]

For x > 45, we have

[tex]Z_{45}=\frac{45-47}{4}=-0.5[/tex]

The required probability becomes

[tex]\begin{gathered} Pr(Z<-2.25)\cup Pr(Z>-0.5) \\ \end{gathered}[/tex]

From the Z-score table/calculator, we have that

[tex]\begin{gathered} Z<-2.25=0.012224 \\ Z>-0.5=0.69146 \end{gathered}[/tex]

So the union sign/or sign means we add. This becomes

[tex]0.012224+0.69146=0.703684[/tex]

Hence the required probability is 0.7037 or 70.37%

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