Question 10·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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Data set $A$ consists of 25 different values that have a minimum of 20, a maximum of 80, a mean of 50, and a standard deviation of 12. The values 20 and 80 are removed from the data set to create data set $B$ . Which choice is true?

For changed data sets, track the mean and spread separately. The mean depends on the total and the count; standard deviation measures spread around the mean. First check whether the removed values balance around the old mean. Then compare their squared distances with the old spread, using Desmos for the arithmetic. Check the mean before comparing distances from it.

Hints

Hint 1
The mean is the total divided by the count. If the removed pair averages to the original mean, it takes away exactly its share of the total. Do $20$ and $80$ balance around $50$ ?
Hint 2
Standard deviation measures spread around the mean. Once you know the mean stays put, the remaining values keep their distances from it. How far is each removed endpoint from $50$ ?
Hint 3
The variance is the average squared distance from the mean. Multiply it by the original count to get the total squared distance, then subtract the contributions of the two removed values.

Step-by-step

Keep the mean, then recalculate the spread
Step 1
Check whether the mean changes
The mean is the total divided by the number of values. The removed values, $20$ and $80$ , balance around the given mean of $50$ , so their own mean is $50$ . Removing two values that average $50$ takes away exactly two values' share of the total. The mean of the remaining values is still $50$ .
Step 2
Measure the removed values' distances
Measure each removed value from the mean that both sets share:
$50-20=30$ $80-50=30$
So both endpoints are $30$ from $50$ . The values left in the set keep their old distances from that mean.
Step 3
Find the original squared-distance total
The standard deviation is found by averaging squared distances from the mean, then taking a square root. To work backward from the given standard deviation of $12$ , type $s=25(12)^2$ in Desmos. It shows $s=3600$ , the total squared distance for the $25$ values in A.
Step 4
Remove the endpoints' contributions
Each endpoint contributed its $30$ -unit distance squared. Type $t=s-2(30)^2$ in Desmos. It shows $t=1800$ , the squared-distance total left in B. This subtraction works because the mean stayed the same, so every remaining value's distance from it stayed the same.
Step 5
Compare the standard deviations
B has $25-2=23$ values. Type $\sqrt{t/23}$ ; Desmos shows about $8.85$ , less than A's $12$ . So B has the same mean, $50$ , and a smaller standard deviation. Choice C.

Question 10·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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