### Statistics Question 7

#### Question 7 - 31 January - Shift 1

If the variance of the frequency distribution

$\mathbf{x} _{\mathbf{i}}$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|---|

Frequency $f_i$ | 3 | 6 | 16 | $\alpha$ | 9 | 5 | 6 |

is 3. Then $\alpha$ is equal to _____.

## Show Answer

#### Answer: 5

#### Solution:

#### Formula: Probability distribution

$\mathbf{f} _{\mathbf{i}}$ | $d_i=$ $\mathbf{x} _i-5$ | $\mathbf{f} _{\mathbf{i}} \mathbf{d} _{\mathbf{i}}^{\mathbf{2}}$ | $\mathbf{f} _{\mathbf{i}} \mathbf{d} _{\mathbf{i}}$ | |
---|---|---|---|---|

2 | 3 | -3 | 27 | -9 |

3 | 6 | -2 | 24 | -12 |

4 | 16 | -1 | 16 | -16 |

5 | $\alpha$ | 0 | 0 | 0 |

6 | 9 | 1 | 9 | 9 |

7 | 5 | 2 | 20 | 10 |

8 | 6 | 3 | 54 | 18 |

$\sigma_x^{2}=\sigma_d^{2}=\frac{\sum f_i d_i{ }^{2}}{\sum f_i}-(\frac{\sum f_i d_i}{\sum f_i})^{2}$

$=\frac{150}{45+\alpha}-0=3$

$\Rightarrow 150=135+3 \alpha$

$\Rightarrow 3 \alpha=15 \Rightarrow \alpha=5$