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Exercise
ABy Admin
May 03'23
Answer
Solution: B
Let [math]X[/math] be normal with mean 10 and variance 4. Let [math]Z[/math] have the standard normal distribution. Let [math]p[/math] = 12th percentile. Then
[[math]]
0.12 = \operatorname{P}(X \leq p ) = \operatorname{P}\left( \frac{X-10}{2} \leq \frac{p-10}{2} \right) = \operatorname{P}( Z \leq \frac{p-10}{2}).
[[/math]]
From the tables, [math]\operatorname{P}( Z ≤ −1.175) =0.12.[/math] Therefore,
[[math]]
\frac{p-10}{2} = -1.175; p-10 = -2.35; p = 7.65.
[[/math]]
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