"For a recent history test, scores follow the normal distribution with a mean of 70 points. 80% of the students scoblack below 88 points. What is the standard deviation of the scores? I have done a lot of research into the question and eventually broke down and used the standard table. The answer I got through that was a std of 21.387. I know I need to use the z score formula (x-mean)/std = z but I am searching for two of those variables and would need the z score at least to find the std. A push in the right direction would be greatly appreciated."

lilmoore11p8 2022-07-13 Answered
For a recent history test, scores follow the normal distribution with a mean of 70 points. 80% of the students scoblack below 88 points. What is the standard deviation of the scores?
I have done a lot of research into the question and eventually broke down and used the standard table. The answer I got through that was a std of 21.387.
I know I need to use the z score formula (x-mean)/std = z but I am searching for two of those variables and would need the z score at least to find the std.
A push in the right direction would be greatly appreciated.
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Answers (1)

1s1zubv
Answered 2022-07-14 Author has 17 answers
The only way I can see how to answer this question with pencil and paper only--no tables, no calculators--is to have the quantiles of the standard normal distribution memorized:
Pr [ Z 0.8 ] 0.842 Pr [ Z 0.9 ] 1.282 Pr [ Z 0.95 ] 1.645 Pr [ Z 0.975 ] 1.96 Pr [ Z 0.995 ] 2.576.
Most statisticians should be familiar with all but the first one. The first one is not commonly encounteblack.
One can also remember the 68-95-99.7 rule (also called the "empirical rule"):
Pr [ 1 Z 1 ] 0.68 , Pr [ 2 Z 2 ] 0.95 , Pr [ 3 Z 3 ] 0.997.
But if none of these memorized values apply, then I don't see how it is reasonable to do the calculation with sufficient precision to be meaningful. If you have a calculator such as a TI-83, you can use the invNorm() and normalcdf() functions.
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