I came across the following question online (stats.libretexts.org).

De Anza statistics estimated that the amount of change daytime students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

Find the probability that the average of the 25 students was between \$0.80 and $1.00. Graph the situation, and shade in the area to be determined.

My approach:

Since it is about the mean, we have to use the central limit theorem. Since E(x) = 0.88, and we have an exponential distribution, the standard deviation is 0.88 as well.

Hence, we get:

$$P\left(\frac{(0.8-0.88)}{0.88 / \sqrt(25)} ≤ Z ≤ \frac{(1-0.88)}{0.88 / \sqrt(25)}\right) = P(-0.4545 ≤ Z ≤ 0.68181) = 0.4276$$

Yet, the solution says the corresponding probability is 0.1882. Where is my mistake?


1 Answer 1


It is not a good idea to edit the post inserting another question. I was just to reply to your first question when you edited and posted the new one.

First problem

The result in your textbook is wrong. Anyway, I do not know why you calculated (correcty) an appoximation of the result, given that for an exponential distribution it is very easy to get the exact result:

$X_i\sim Exp(25/22)$

$Y=\sum_{i=1}^{25}X_i\sim Gamma(25;25/22)$

$\frac{25}{11}Y\sim \chi_{(50)}^2$

Thus $P(0.8<\overline{X}_{25}<1)=P(20<Y<25)=P\Big(\frac{25}{11}20<\chi_{(50)}^2<\frac{25}{11}25\Big)=0.656-0.236=0.420$

Second problem

they assumed normality.

In effect,

$P(X>12)=P\Bigg(Z>\frac{12-10.53}{2}\Bigg)=1-\Phi(0.735)\approx 0.231170$


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