- #1

k77i

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## Homework Statement

Suppose that X is normally distributed with mean 95 and standard deviation 17.

A. What is the probability that is greater than 126.79?

B. What value of X does only the top 16% exceed?

## Homework Equations

z= (X-(mean of x))/standard deviation

## The Attempt at a Solution

using the equation, I found z= (126.79-95)/17=1.87

then using a statistical table, the value for P(z greater than 1.87) = 1 - P(z=1.87) = 0.0307

But this is wrong..

For B)

I considered that 84% must be the value which the top 16% must exceed.. then 84% of 95 is 88.42

Also wrong (well I wasn't really too sure about part B)

Any help would be appreciated!