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A company pays it employees an average wage of $ \$15.90$ an hour with standard deviation $\$1.50$ per hour. Assume the wages are approximately normally distributed. What is the hourly wage that is exceeded by only $5\%$ of the employees?

Is the setup $P(X \geq x) = 0.05$ correct? Also, standardizing to $P(Z \geq \frac{x-15.9}{1.5}) = 0.05$

My final answer was $x = 18.375$

Did I set it up correctly?

Lemon
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    The setup is almost correct, it said exceeded so it should be $\Pr(X\gt x)=0.05$. Of course it doesn't make any difference, since this is a continuous distribution, but one has to get used to being sensitive to the exact words, since it can make a big difference in a discrete distribution. By table, by interpolation, gives about $1.645\sigma$ for the place that has $5%$ in the right tail, so the answer is closer to $18.37$, a bit under. – André Nicolas Jul 30 '13 at 06:26
  • The information has been conveyed! I hesitate to post an answer that basically only says, yes, you are right. – André Nicolas Jul 30 '13 at 06:41
  • A +1 then. I can't leave you empty-handed – Lemon Jul 30 '13 at 06:43

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