Question

A grocery store has an express line for customers purchasing at most five items. Let \(x\) be the number of items purchased by a randomly selected customer using this line. Give examples of two different assignments of probabilities such that the resulting distributions have the same mean but quite different standard deviations.

Short answer

Two assignments could be: 1) Uniform distribution with probabilities 1/6, resulting in mean 2.5 and standard deviation of approximately 1.707 and 2) Non-uniform assignment with probabilities such that P(X=0) = P(X=5) = 0.4 and P(X=1) = P(X=4) = 0.1 with mean 2.5 and standard deviation of approximately 2.121

Step-by-step solution

- Understanding Mean and Standard Deviation

The mean (or expected value) of a discrete random variable \(X\) is given by \(\sum{xP(X=x)}\). The standard deviation, reflecting the spread of the probability distribution, is calculated by \(\sqrt{\sum{(x-\mu)^2 P(X=x)}}\), where \(\mu\) is the mean.

- Constructing the first assignment

Let's consider a uniform assignment such that \(P(X=x)\) is \(1/6\) for \(0 \leq x \leq 5\). The mean is then \(\sum{xP(X=x)} =\sum_{x=0}^{5}x(1/6) = 2.5\). The variance is given by \(\sum{(x-2.5)^2 P(X=x)} = (-2.5)^2(1/6)+ (-1.5)^2(1/6)+(-0.5)^2(1/6) + (0.5)^2(1/6)+(1.5)^2(1/6)+(2.5)^2(1/6) = 2.91667\), giving a standard deviation of \(\sqrt{2.91667}=1.7078251\).

- Constructing the second assignment

Let's consider a non-uniform assignment, such as \(P(X=0) = P(X=5) = 0.4\), \(P(X=1) = P(X=4) = 0.1\) and \(P(X=2) = P(X=3) = 0\). The mean is same as before: \(\sum{xP(X=x)} = 2.5\). The variance is calculated as \(\sum{(x-2.5)^2 P(X=x)} = (-2.5)^2(0.4)+ (-1.5)^2(0.1)+(-0.5)^2(0)+(0.5)^2(0)+(1.5)^2(0.1)+(2.5)^2(0.4) = 4.5\). The standard deviation, thereby is \(\sqrt{4.5}=2.1213203\) which is indeed greater than before.