## Sunday, 30 October 2011

### Statistics-1 (XAT-2011) (5-Marks)

In a list of seven integers, one integer denoted as x is unknown. The other six integers are 20,4,10,4,8 and 4. If the mean, median and mode of these 7 integers are arranged in increasing order, they form an arithmetic progression. The sum of all possible values of x is:
(1)26   (2) 32   (3) 34   (4) 38   (5) 40
Solution:
Arranging the given six integers in increasing order:
4,4,4,8,10,20
Irrespective of the value of x, the number of occurrences being three, 4 is the mode.

Mode = 4; Mean = (4+4+4+8+10+20+x)/7 = (50+x)/7

Here three cases arise:
Case-I                 If x≤4, median = 4
Case-II                If 4
Case-III              If x>8, median = 8

Case-I:
x≤4;       Median = 4;
Given that mean, median and mode arranged in ascending are in AP
Since mode=median=4, ascending order is not possible here and hence this case is excluded.
Case-II:
4
4 possible values for x are 5,6,7 and 8
But, Mean which is (50+x)/7 is integer only for x=6
=> x = 6
Case-III:
x>8;      Median = 8;
x>8 => (50+x)/7 > 8
Hence 4, 8, (50+x)/7 are in AP => 8 - 4 = (50+x)/7 - 8 => 4 = (50+x-56)/7
=> 28 = x-6 => x = 34

In all the cases, the possible values of x are 6 and 34
Sum of possible values of x = 6+34 = 40