an anthropologist wishes to estimate the average height of men for a certain race of people. if the population standard deviation is assumed to be 2.5 inches and if she randomly samples 100 mean, find the probability that the difference between the sample mean and the true population mean will not exceed 0.5 inch.

  • $\begingroup$ Hint: Imagine taking a sample, let their heights be $X_1$ to $X_{100}$, and let $Y=\frac{\sum X_i}{100}$. Then $Y$ is roughly normal, mean the population mean, standard deviation $\frac{2.5}{\sqrt{100}}$. $\endgroup$ Apr 16, 2014 at 20:45
  • $\begingroup$ im not really sure how to calculate this if we dont have the value of population mean. $\endgroup$ Apr 16, 2014 at 21:05
  • $\begingroup$ It asks about the difference between sample mean and population mean. That difference has mean $0$. The one thing that is ambiguous is what they mean by difference. Is it ordinary difference or absolute value of difference? $\endgroup$ Apr 16, 2014 at 21:12
  • $\begingroup$ im thinking that they mean absolute value of difference . since they gave a 0.5 inch value. $\endgroup$ Apr 16, 2014 at 21:18
  • $\begingroup$ Possibly. I wish they were less vague. Anyway, it then comes down to finding the probability that $|Z|\le \frac{0.5}{2.5/10}$. Now use normal tables, or software. $\endgroup$ Apr 16, 2014 at 21:26

1 Answer 1


We will assume that by "difference does not exceed $0.5$ inch," what is meant is that the absolute value of the difference does not exceed $0.5$ inch.

Let the heights of the people in the sample be $X_1,X_2,\dots,X_{100}$. Then the sample mean is the random variable $Y$, where $$Y=\frac{1}{100}(X_1+X_2+\cdots+X_{100}).$$ Let the population mean be $mu$. Then $Y$ has mean $\mu$ and standard deviation $\frac{2.5}{\sqrt{100}}$.

We assume that the distribution of the $X_i$ is reasonably nice. Then $Y$ has a close to normal distribution. From here on, we assume the distribution is normal. Then $Y-\mu$ is normal with mean $0$ and standard deviation $0.25$. Then $$\Pr\left(|Y-\mu|\le 0.5\right)=\Pr\left(|Z|\le \frac{0.5}{0.25}\right).$$ This can be found using tables of the standard normal, or software.


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