Q.
Consider the following data which represents the runs scored by two batsmen in their last ten matches as
Batsman A : $30,91,0,64,42,80,30,5,117,71$
Batsman B : $53,46,48,50,53,53,58,60,57,52$
Which of the following is/are true about the data?
I. Mean of batsman $A$ runs is $53$ .
II. Median of batsman $A$ runs is $42$ .
III. Mean of batsman $B$ runs is $53$ .
IV. Median of batsman $B$ runs is $53$ .
Statistics
Solution:
The runs scored by two batsmen in their last ten matches are as follows Batsman A: $30,91,0,64,42,80,30,5,117,71$
Batsman B :$ 53,46,48,50,53,53,58,60,57,52$
Clearly, the mean and median of the data are
Batsman A
Batsman B
Mean
53
53
Median
53
53
We calculate the mean of a data (denoted by $\bar{x}$),
i.e., $\bar{x} = \frac{1}{n} \displaystyle\sum x_i$
Also, the median is obtained by first arranging the data in ascending or descending order and applying the rules
Mean for batsman $A$
$ = \frac{30 + 91+ 0 + 64+42 + 80 + 30 + 5 + 117 + 71}{10}$
$ = \frac{530}{10} = 53$
Mean for batsman $B$
$ = \frac{53 +46 + 48 +50+53+ 53 + 58 +60 +57 + 52}{10}$
$ = \frac{530}{10} = 53$
To apply the formula to obtain median first arrange the data in ascending order
For batsman A
0
5
30
30
42
64
71
80
91
117
For batsman A
46
48
50
52
53
53
53
57
58
60
Here, we have $n = 10$ which is even number. So median is the mean of $5^{th}$ and $6^{th}$ observations.
Median for batsman $A = \frac{42 + 64}{2}$
$ = \frac{106}{2} = 53$
Median for batsman $B = \frac{53 + 53}{2}$
$ = \frac{406}{2} = 53$
| Batsman A | Batsman B | |
|---|---|---|
| Mean | 53 | 53 |
| Median | 53 | 53 |
| For batsman A | 0 | 5 | 30 | 30 | 42 | 64 | 71 | 80 | 91 | 117 |
| For batsman A | 46 | 48 | 50 | 52 | 53 | 53 | 53 | 57 | 58 | 60 |