Question

A data set of 27 different numbers has a mean of 33 and a median of 33. A new data set is

Answer 1

Answer:

Option D.

Step-by-step explanation:

Suppose that we have a set of N values:

{x₁, x₂, ..., xₙ}

The median is the value in the middle, and the mean is calculated as:

$M = \frac{(x_1 + x_2 + x_3 ... + x_n)}{N}$

Because here we have "the median" then we will assume that N is odd, and there is only one median, the value "k"  (if N was even, the median would be the mean of the two middle values, that case is really similar to the case where N is odd, so solving only one of the cases is enough)

Now we add 7 to all the values greater than the median

We subtract 7 to all values smaller than the median.

Then the median remains unchanged (because we did not add nor subtract anything to the median).

The new mean will be:

$M' = \frac{(x_1 - 7) + (x_2 - 7) + ... + (x_{k}) + ... + (x_n + 7)}{N} = \frac{(x_1 + x_2 + ... + x_n) + (-7)*(n/2 - 0.5) + 7*(n/2 - 0.5)}{N} = \frac{(x_1 + x_2 + ... + x_n)}{N} = M$

So the mean does not change.

Because the mean is computed as the sum of the numbers divided by N, and the mean does not change, and N does not change, then the sum of the numbers does not change.

Finally, the standard deviation is computed as:

$SD = \sqrt{\frac{(x_1 - M)^2 + ... + (xn - M)^2}{N} }$

Because M does not change, if we add or subtract numbers to some of the values, the standard deviation will change (Because all the terms are squared, so the added and subtracted sevens don't cancel like in the previous cases)

Then the only value that does not have the same value in both the original and new data sets is the standard deviation.