Given: mu = 65.7 sigma = 15 Standard normal distribution Question: 20% of the area will be less than what x value? I looked at my z-score table and couldn't find .0200 so now I have no clue how to even begin!
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From your z-score table $\displaystyle P(z< -0.8416)= .2$ Use this fact in $\displaystyle z = \frac{x-\mu}{\sigma} \implies -0.8416 = \frac{x-65.7}{15}$ to solve for $\displaystyle x$
Where did you get .8416? Could you possibly have a more complete z-score table or am I missing some intermediary calculation?
Originally Posted by tom ato Where did you get .8416? Could you possibly have a more complete z-score table or am I missing some intermediary calculation? You could just as easily use -0.84. There is no missing step here, and you no doubt are meant to use -0.84.
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