Suppose 20% of the people in a city prefer Pepsi-Cola as their soft drink of choice. If a random sample of six people is chosen, the number of Pepsi drinkers could range from zero to six. Shown here are the possible numbers of Pepsi drinkers in a sample of six people and the probability of that number of Pepsi drinkers occurring in the sample. Use the data to determine the mean number of Pepsi drinkers in a sample of six people in the city, and compute the standard deviation.

Number of Pepsi Drinkers | Probability |
---|---|

0 | 0.262 |

1 | 0.393 |

2 | 0.246 |

3 | 0.082 |

4 | 0.015 |

5 | 0.002 |

6 | 0.000 |

#### Solution

Let $X$ denote number of Pepsi drinkers in a sample of six people in the city.

The mean number of Pepsi drinkers in a sample of six people in the city is

` $$ \begin{aligned} E(X) & = \sum_x x*P(X=x)\\ & = (0*0.262)+(1*0.393)\\ & \quad +(2*0.246)+(3*0.082)\\ &\quad +(4*0.015)+(5*0.002)+(6*0)\\ & = 1.201 \end{aligned} $$ `

` $$ \begin{aligned} E(X^2) & = \sum_x x^2*P(X=x)\\ & = (0*0.262)+(1*0.393)\\ & \quad +(4*0.246)+(9*0.082)\\ &\quad +(16*0.015)+(25*0.002)+(36*0)\\ & = 2.405 \end{aligned} $$ `

The variance of number of Pepsi drinkers in a sample of six people in the city is

` $$ \begin{aligned} V(X) & = E(X^2) - [E(X)]^2\\ & = 2.405 - [1.201]^2\\ & = 0.9626 \end{aligned} $$ `

The standard deviation is $sd(X) = \sqrt{V(X)} = 0.9811$.