For a recent history test, scores follow the normal distribution with a mean of 70 points. 80% of the students scored 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.
Best Answer
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:
$$\begin{align*} \Pr[Z \le 0.8] &\approx 0.842 \\ \Pr[Z \le 0.9] &\approx 1.282 \\ \Pr[Z \le 0.95] &\approx 1.645 \\ \Pr[Z \le 0.975] &\approx 1.96 \\ \Pr[Z \le 0.995] &\approx 2.576. \end{align*}$$ Most statisticians should be familiar with all but the first one. The first one is not commonly encountered.
One can also remember the 68-95-99.7 rule (also called the "empirical rule"): $$\begin{align*} \Pr[-1 \le Z \le 1] &\approx 0.68, \\ \Pr[-2 \le Z \le 2] &\approx 0.95, \\ \Pr[-3 \le Z \le 3] &\approx 0.997.\end{align*}$$
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()andnormalcdf()functions.