**Question 1**:

Table shows the age of 40 tourists who visited a tourist spot.

Age |
Tourists |

10 – 19 |
4 |

20 – 29 |
m |

30 – 39 |
n |

40 – 49 |
10 |

50 – 59 |
8 |

Given that the median age is 35.5, find the value of

*m*and of*n*.

*Solution:*Given that the median age is 35.5, find the value of

*m*and of*n*.Age |
Frequency |
Cumulative frequency |

10 – 19 |
4 |
4 |

20 – 29 |
m |
4 + m |

30 – 39 |
n |
4 + m + n |

40 – 49 |
10 |
14 + m + n |

50 – 59 |
8 |
22 + m + n |

22 +

*m*+*n*= 40*n*= 18 –

*m*—–(1)

Given median age = 35.5, therefore median class = 30 – 39

$\begin{array}{l}35.5=29.5+\left(\frac{20-\left(4+m\right)}{n}\right)\times 10\\ 6=\left(\frac{16-m}{n}\right)\times 10\end{array}$

6

*n*= 160 – 10*m*3

*n*= 80 – 5*m*—–(2)Substitute (1) into (2).

3 (18 –

*m*) = 80 – 5*m*54 – 3

*m*= 80 – 5*m*2

*m*= 26*m*= 13

Substitute

*m*= 13 into (1).*n*= 18 – 13

*n*= 5

**Thus**

*m*= 13,*n*= 5.**Question 2**:

A set of examination marks

*x*_{1},*x*_{2},*x*_{3},*x*_{4},*x*_{5},*x*_{6}has a mean of 6 and a standard deviation of 2.4.**(a)**Find

**(i)**the sum of the marks, $\Sigma x$ ,

**(ii)**the sum of the squares of the marks, $\Sigma {x}^{2}$ .

**(b)**Each mark is multiplied by 2 and then 3 is added to it.

Find, for the new set of marks,

**(i)**the mean,

**(ii)**the variance.

*Solution:***(a)(i)**

$\begin{array}{l}\text{Givenmean}=5\\ \frac{\Sigma x}{6}=6\\ \Sigma x=36\end{array}$

**(a)(ii)**

$\begin{array}{l}\text{Given}\sigma =2.4\\ {\sigma}^{2}={2.4}^{2}\\ \frac{\Sigma {x}^{2}}{n}-{\overline{X}}^{2}=5.76\\ \frac{\Sigma {x}^{2}}{6}-{6}^{2}=5.76\\ \frac{\Sigma {x}^{2}}{6}=41.76\\ \Sigma {x}^{2}=250.56\end{array}$

**(b)(i)**

Mean of the new set of numbers

= 6(2) + 3

= 15

**(b)(ii)**

Variance of the original set of numbers

= 2.4

^{2 }= 5.76Variance of the new set of numbers

= 2

^{2}(5.76)= 23.04