Question

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 $53$ .

• A. Only I is true
• B. I and III are true
• C. I, III and IV are true
• D. All are true

Answer 1

Answer: (C)

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., $\bar{x}=\frac{1}{n}\sum x_i$
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 $B=\frac{53+53}{2}$
$=\frac{406}{2}=53$