Statistics and Dispersion

Standard deviation will remain the same.

The sum of deviations will become zero when the deviations are taken from their mean.

Mean gets increased by 4.

Range = 200 standard deviation = 100

Quartile deviation is half of inter-quartile range.

Class mark is the mid value of a class.

The sum of the first 200 natural numbers = (200 * 201)/2 = 20100 avg. = 20100/200 = 100.5

(4 * 25 + 4 * 30 + 4 * 35)/12 = 30

Every number of the second set would be 12 more than its countrpart. So, average will be 12 more than m.

The answer is 50 * 202/100 = 101

Sum of first three numbers = 141 Sum of last three numbers = 174 Since the third number is common in both, so difference will be 33.

Whole numbers start from 0; hence n whole numbers means n-1 natural numbers. Sum first n-1 natural numbers is {(n-1)(n-1+1)}/2 ; Average is ((n-1)*n)/(2*n) , Average is (n-1)/2.

Average of consecutive integers is always middle term. So, the average of first 129 terms is (129 + 1)/2= 65^th term, which is n. Average of 137 terms is (137 + 1)/2 = 69^th term. So, 69^th term is 65^th term + 2*4 = n + 8.

Since the average is 8 more than the actual, the second number will increase by 72 (9 x 8) by interchanging the digits. If 'ab' is the second number, then 10a + b + 72 = 10b+ a, i.e. 9 (b - a) = 72b - a = 8 The possible number 'ab' is 19. Since the second number is 19, the first number is 18. So, X = 18

Sum of 5 consecutive odd numbers (avg. m) = 5m Sum of 7 consecutive odd numbers (avg. n) = 7n So, resultant avg. = (5m + 7n)/12