Statistics Long Questions (Question 1 & 2)


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

35.5=29.5+( 20( 4+m ) n )×10 6=( 16m 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, Σx ,
     (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 Σx 6 =6 Σx=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