Question

are my answers wrong?

Find the value of * that yields the probability shown, where X is a normally distributed random variable X with mean 54 and standard deviation 12. 1.P/X<x*)-0.0900 37.9109 2.PIX>z*)-0.6500 58.6238

Answer 1

Solution

given : Mean (u) = 54, Standard deviation (0) = 12

formula : Z= X-H

1. P(X<r*) = 0.0900

X-" < →P r* - 54 12 0.0900

→P Z< .-* - 54 12 -0.0900

Refer Standard normal table/Z-table, Lookup for z-score corresponding to area 0.0900 to the left of the normal curve OR use excel formula "=NORM.S.INV(0.0900)" to find the z-score.

$\small \Rightarrow P(Z<\mathbf{-1.34076 } )=0.0900$

$\small -1.34076 =\frac{x^{*}-54}{12}$

$\small x^{*}=54+\left (-1.34076 *12 \right )$

$\small x^{*}=54-16.0891$

$\small x^{*}=\mathbf{37.9109}$

$\small {\color{Red} {\color{Red} }\mathbf{37.9109}}$

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$\small {\color{blue} \mathbf{2.}}\;P(X>x^{*})=0.6500$

$\small \Rightarrow P\left ( \frac{X-\mu}{\sigma}>\frac{x^{*}-54}{12} \right )=0.6500$

$\small \Rightarrow P\left (Z>\frac{x^{*}-54}{12}\right )=0.6500$

Refer Standard normal table/Z-table, Lookup for z-score corresponding to area 0.6500 to the right of the normal curve OR use excel formula "=NORM.S.INV(1-0.6500)" to find the z-score.

$\small \Rightarrow P(Z>\mathbf{-0.38532 } )=0.6500$

$\small -0.38532 =\frac{x^{*}-54}{12}$

$\small x^{*}=54+\left (-0.38532 *12 \right )$

$\small x^{*}=54-4.6238$

$\small x^{*}=\mathbf{49.3762 }$

$\small {\color{Red} {\color{Red} }\mathbf{49.3762 }}$