# Normal distribution percentage problem.

## Homework Statement

A company pays its employees an average wage of \$3.25 an hour with a standard deviation of sixty cents. If the wages are approximately normally distributed, determine
the minimum wage of the employees who are paid the highest 5%.

z=(x-μ)/σ

## The Attempt at a Solution

I did:
.95=((x-3.25)/.6) which equals .57=x-3.25
therefore x=3.82. I feel like I didn't do this correctly and I can't finish the rest of my homework if I don't know what I'm doing. Thanks.

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Isn't that equation for standardizing your values? 95% has nothing to do with z-scores.

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Isn't that equation for standardizing your values? 95% has nothing to do with z-scores.
Oh, shoot, you're right. Well how would I find out what 95% of the wages cause I don't have the slightest clue.

Your z-value is +-σ from the mean. What is 95% in terms of σ on a normal distribution? Also, do you use calculators to help you in your class? If you do, this question should be quite straightforward.

Ray Vickson
Homework Helper
Dearly Missed
Oh, shoot, you're right. Well how would I find out what 95% of the wages cause I don't have the slightest clue.
If z_95 is the 95th percentile of the standard normal (which is available in tables or on some calculators) you need $$z_{95} = \frac{x-3.25}{0.6}.$$ Note: you should divide by 0.6 not 6, since 0.6 is the standard deviation in dollars.

RGV

Your z-value is +-σ from the mean. What is 95% in terms of σ on a normal distribution? Also, do you use calculators to help you in your class? If you do, this question should be quite straightforward.
No, my teacher wants us only using this table she gave us.