Consider statement I. When every number in set T is multiplied by 3, the average is multiplied by 3. The average of the numbers in set T must be `3(14)=42`. Statement I must be true. Eliminate (B), which does not contain I.
You can also show that statement I is true algebraically. If the members of the set S are `x_(1), x_(2), x_(3),...,x_(n)`, then the average of the numbers in set S is `(x_(1)+x_(2)+...+x_(n))/(n)`. The members of set T are `3x_(1), 3x_(2), 3x_(3),...,3x_(n)`. The average of the numbers in set T is `(3x_(1)+3x_(2)+...+3x_(n))/(n)=(3(x_(1)+x_(2)+....+x_(n)))/(n)= 3((x_(1)+x_(2)+....+x_(n))/(n))`. The average 3 `3((x_(1)+x_(2)+....+x_(n))/(n))` of the numbers in set T is 3 times the average `(x_(1)+x_(2)+...+x_(n))/(n)` of the numbers in set S. Thus, the average of the numbers in set T is 3(14) = 42.
Now consider statement II. Suppose that the members of set S, in increasing order, are `y_(1),y_(2),y_(3),...,y_(n)`. Then the members of set T, in increasing order, are `3y_(1), 3y_(2), 3y_(3),...., 3y_(n)`. If there is an odd number of numbers in set S - that is, if n is odd - then the middle term among `y_(1), y_(2), y_(3),...., y_(n)`, which is the middle term among `y_(1), y_(2), y_(3),....y_(n)`, which is the median of set S, was multiplied by 3 and is the middle term among `3y_(1), 3y_(2), 3y_(3), ....,3y_(n)`, which is the median of set T. If there is an even number of numbers in set S, then the median of the numbers in set S is the average of the two middle terms. Since the two middle terms among `3y_(1), 3y_(2), 3y_(3),....,3y_(n)` are the two corresponding middle terms among `y_(1), y_(2), y_(3), ....,y_(n)` that were multiplied by 3, the average of the two middle terms among `3y_(1), 3y_(2), 3y_(3),...,3y_(n)` must be the average of the two middle terms among `y_(1), y_(2), y_(3), ...., y_(n)` multiplied by 3. Thus, the median of set T must be 3(12) = 36. Statement II must be true, so eliminate (A) and (D) which do not contain II.
Consider statement III. If each number is multiplied by 3, the numbers are 3 times more dispersed. So the standard deviation of the numbers in set T must be 3 times the standard deviation of the numbers in set S. The standard deviation of the numbers in set T is 3(1.8) = 5.4. Statement III must be true. Statements I, II, and III must all be true, and (E ) is correct.
You can also show that statement III is true algebraically. If the numbers in set S are `x_(1), x_(2), x_(3),...,x_(n)` then the standard deviation of the numbers in set S is `(sqrt((x_(1)-bar(x))^(2)+(x_(2)-bar(x))^(2)+...+(x_(n)-bar(x))^(2)))/(n)`
where `bar(x)` is the average of the numbers in set S. Thus, `bar(x)=(x_(1)+x_(2)+...+x_(n))/(n)`. You saw when considering statement I that the average of the numbers in set T is `3bar(x)`, where `bar(x)` is the average of the numbers in set S. The standard deviation of the numbers in set T is
`(sqrt((3x_(1)-3bar(x))^(2)+(3x_(2)-3bar(x))^(2)+...+(3x_(n)-3bar(x))^(2)))/(n)`
`= (sqrt([3(x_(1)-bar(x))]^(2)+[3(x_(2)-bar(x))]^(2)+...+[3(x_(n)-bar(x))]^(2)))/(n)`
`= (sqrt(9(x_(1)-bar(x))^(2)+9(x_(2)-bar(x))^(2)+...+9(x_(n)-bar(x))^(2)))/(n)`
`=sqrt(9(((x_(1)-bar(x))^(2)+(x_(2)-bar(x))^(2)+...+(x_(n)-bar(x))^(2))/(n)))`
`= sqrt(9) (sqrt((x_(1)-bar(x))^(2)+(x_(2)-bar(x))^(2)+...+(x_(n)-bar(x))^(2)))/(n)`
`= 3((sqrt((x_(1)-bar(x))^(2)+(x_(2)-bar(x))^(2)+...+(x_(n)-bar(x))^(2)))/(n))`
So the standard deviation of the numbers in set T,
`= 3((sqrt((x_(1)-bar(x))^(2)+(x_(2)-bar(x))^(2)+...+(x_(n)-bar(x))^(2)))/(n))` , is 3
times the standard deviation of the numbers in set
`S, (sqrt((x_(1)-bar(x))^(2)+(x_(2)-bar(x))^(2)+...+(x_(n)-bar(x))^(2)))/(n)`
