parkhydel wrote:
Consider the sets S, T and U, where
S = {31, 14, 64, 22, 43, 66},
T = {x, 31, 14, 64, 22, 43, 66}, and
U = {y, 31, 14, 64, 22, 43, 66}
The mean of T is 5 less than the mean of S. The median of U is 4 less than the median of S.
Select the value of x and the value of y consistent with the statements given. Make two selections, one in each column.
You should observe that all the values in S are also there in other sets, so you just require to calculate mean once.
sayan640, you have a calculator to ease your job. But if you wanted to do the calculations by yourself, S = {31, 14, 64, 22, 43, 66}: the units digit (4,4,2) and (1,3,6) will give you 10 each, so total 20. Ten's digit = 3+1+6+2+4+6 = 22.....Total = 220+20 = 240 and mean = 240/6 = 40
Now, average of T is 5 less.
x would be 40 if the average remained same. But 5 less means each of the 7 elements in T are to be reduced by 5, and this reduction is catered by x, so x= 40-5*7 = 5.
OR
Average of T = \(\frac{x+ 31+14+64+ 22+ 43+ 66}{7} = 35......\frac{x+240}{7}=35......x=245-240=5\)
Next, let us check on Median
Median of S = average of middle two values as there are even number of elements = \(\frac{31+43}{2 } = 37\)
So Median of U = 37-4 = 33.
As U has odd number of elements, the median should be an element,
but there is no 34 in the given values. So, y=33