Question

The probability distribution of the daily number of customers at the YumYum is given below. Daily number of customers Probability 20 0.2 40 0.3 60 0.4 80 0.1 What is the standard deviation of the number of daily customers

Answer 1

18.33

Explanation:

Given the probability distribution below:

`\left|\begin{array}cccc$Daily number of customers& 20 & 40 &60 & 80 \\$Probability&0.2&0.3&0.4 &0.1\end{array}\right|`

We are required to determine the standard deviation of the number of daily customers.

`\text{Standard Deviation}=√(\sum (x-E(x))^2 \cdot P(x))`

Mean, E(x)=20*0.2+40*0.3+60*0.4+80*0.1=48

`\left|\begin{array}cx&x-E(x)&(x-E(x))^2&P(x)&(x-E(x))^2 \cdot P(x)\\20&-28&784&0.2&156.8\\40&-8&64&0.3&19.2\\60&12&144&0.4&57.6\\80&32&1024&0.1&102.4\\Total&&&&336\end{array}\right]`

Standard Deviation
`=√(336)=18.33`