lilmoore11p8

2022-07-13

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.

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.

1s1zubv

Beginner2022-07-14Added 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:

$\begin{array}{rl}Pr[Z\le 0.8]& \approx 0.842\\ Pr[Z\le 0.9]& \approx 1.282\\ Pr[Z\le 0.95]& \approx 1.645\\ Pr[Z\le 0.975]& \approx 1.96\\ Pr[Z\le 0.995]& \approx \mathrm{2.576.}\end{array}$

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"):

$\begin{array}{rl}Pr[-1\le Z\le 1]& \approx 0.68,\\ Pr[-2\le Z\le 2]& \approx 0.95,\\ Pr[-3\le Z\le 3]& \approx \mathrm{0.997.}\end{array}$

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.

$\begin{array}{rl}Pr[Z\le 0.8]& \approx 0.842\\ Pr[Z\le 0.9]& \approx 1.282\\ Pr[Z\le 0.95]& \approx 1.645\\ Pr[Z\le 0.975]& \approx 1.96\\ Pr[Z\le 0.995]& \approx \mathrm{2.576.}\end{array}$

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"):

$\begin{array}{rl}Pr[-1\le Z\le 1]& \approx 0.68,\\ Pr[-2\le Z\le 2]& \approx 0.95,\\ Pr[-3\le Z\le 3]& \approx \mathrm{0.997.}\end{array}$

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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