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Question

Your company, Sonic Video, Inc., has conducted research that shows the following probability distribution, where X is the number of video arcades in a randomly chosen city with more than 500,000 inhabitants:

x0123456789P(X=x).07.09.35.25.15.03.02.02.01.01\begin{matrix} \text{x} & \text{0} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & \text{7} & \text{8} & \text{9}\\ \text{P(X=x)} & \text{.07} & \text{.09} & \text{.35} & \text{.25} & \text{.15} & \text{.03} & \text{.02} & \text{.02} & \text{.01} & \text{.01}\\ \end{matrix}

a. Compute the mean, variance, and standard deviation (accurate to one decimal place). b. As CEO of Startrooper Video Unlimited, you wish to install a chain of video arcades in Sleepy City, U.S.A. The city council regulations require that the number of arcades be within the range shared by at least 75% of all cities. What is this range? What is the largest number of video arcades you should install so as to comply with this regulation?

Solution

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Answered 2 years ago
Answered 2 years ago
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a.

For a finite random variable XX taking on values x1,x2,,xnx_{1},x_{2},\ldots,x_{n},

the variance of XX is

σ2=E([Xμ]2)=(x1μ)2P(X=x1)+(x2μ)2P(X=x2)++(xnμ)2P(X=xn)=i(xiμ)2P(X=xi).\begin{align*} \sigma^{2}& =E([X-\mu]^{2}) \\ &=(x_{1}-\mu)^{2}P(X=x_{1})+(x_{2}-\mu)^{2}P(X=x_{2})+\cdots+(x_{n}-\mu)^{2}P(X=x_{n}) \\ & =\displaystyle \sum_{i}(x_{i}-\mu)^{2}P(X=x_{i}) . \end{align*}

The standard deviation of XX is then the square root of the variance:

σ=σ2.\sigma=\sqrt{\sigma^{2}}.

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