Question 7·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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Data set $A$ consists of 12 values and has mean $m$ and standard deviation $s$ . Two of the values in data set $A$ are 4 and 22. Data set $B$ is created by replacing 4 and 22 with 10 and 16.

Which choice best compares the mean and standard deviation of data set $B$ with those of data set $A$ ?

For a replacement of a few data values, check the mean and standard deviation separately. Compare the old and new totals to settle the mean. Then compare squared distances from the mean to settle the spread; Desmos can handle the arithmetic. A shared midpoint for the replaced pairs is useful, but it need not be the mean of the whole data set.

Hints

Hint 1
The mean is the total divided by the number of values. Both sets still have 12 values, so compare the sum of the two values removed with the sum of the two replacements.
Hint 2
Each changed pair has a midpoint, the number halfway between its values. That midpoint might not be the mean of all 12 values. How far is each member of each pair from the midpoint?
Hint 3
Standard deviation depends on squared distances from the whole set's mean. Write that mean as $m$ , and compare the two pairs' squared distances from $m$ . The unknown-mean terms may cancel.

Step-by-step

Compare totals, then squared distances
Step 1
Check whether the mean changes
Type $4+22$ and $10+16$ in Desmos. Both show $26$ . The replaced values have the same total as the values they replace, and both sets contain 12 values. Since the mean is the total divided by the count, data set $B$ also has mean $m$ .
Step 2
Find the changed pairs' common midpoint
Add $26/2$ , $13-4$ , and $13-10$ in Desmos. They show $13$ , $9$ , and $3$ . So the old pair is $13-9,13+9$ , while the new pair is $13-3,13+3$ . Their midpoint is $13$ , but don't assume $m=13$ : the other 10 values affect the whole set's mean.
Step 3
Compare squared distances for any mean
Let $d=m-13$ , the distance from the pairs' midpoint to the unknown mean, with a sign showing which side $m$ is on. For a pair $13-a,13+a$ , square its distances from $m$ :
$(d+a)^2+(d-a)^2$
Expand each square:
$(d^2+2da+a^2)+(d^2-2da+a^2)$
Cancel the opposite middle terms:
$2d^2+2a^2$
Use the offsets from the previous step. If $S_A$ and $S_B$ are the changed pairs' sums of squared distances, then:
$S_A=2d^2+2(9^2),\qquad S_B=2d^2+2(3^2).$
This works even when $m$ isn't $13$ .
Step 4
Use the smaller squared-distance total
Subtract, canceling the equal $2d^2$ terms:
$S_A-S_B=2(9^2-3^2).$
Type $2(9^2-3^2)$ in Desmos; it shows $144$ . The other 10 values have unchanged squared distances from the unchanged mean. With the same count and mean, a smaller total of squared distances means a smaller standard deviation. So data set $B$ has mean $m$ and standard deviation less than $s$ . Choice C.

Question 7·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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