7.5.1 Statistics Long Questions (Question 1 & 2)

Question 1:

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

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.

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 + (\frac{20 - (4 + m)}{n}) \times 10 \\ 6 = (\frac{16 - m}{n}) \times 10 \end{array}

6n = 160 – 10m

3n = 80 – 5m —–(2)

Substitute (1) into (2).

3 (18 – m) = 80 – 5m

54 – 3m = 80 – 5m

2m = 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 x₁, x₂, x₃, x₄, x₅, x₆ has a mean of 6 and a standard deviation of 2.4.

(a) Find

(i) the sum of the marks,

Σ x

(ii) the sum of the squares of the marks,

Σ 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} Given mean = 5 \\ \frac{Σ x}{6} = 6 \\ Σ x = 36 \end{array}

(a)(ii)

\begin{array}{l} Given σ = 2.4 \\ σ^{2} = {2.4}^{2} \\ \frac{Σ x^{2}}{n} - {\bar{X}}^{2} = 5.76 \\ \frac{Σ x^{2}}{6} - 6^{2} = 5.76 \\ \frac{Σ x^{2}}{6} = 41.76 \\ Σ 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

= 2.4²= 5.76

Variance of the new set of numbers

= 2² (5.76)

= 23.04