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

35.5 = 29.5 + ( 20 ( 4 + m ) n ) × 10 6 = ( 16 m n ) × 10
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)
Given mean = 5 Σ x 6 = 6 Σ x = 36

(a)(ii)
Given  σ = 2.4 σ 2 = 2.4 2 Σ x 2 n X ¯ 2 = 5.76 Σ x 2 6 6 2 = 5.76 Σ x 2 6 = 41.76 Σ x 2 = 250.56

(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