Question 54·Hard·One-Variable Data Distributions; Measures of Center and Spread
Two data sets of 24 integers each are summarized in the histograms shown.
For each histogram, each interval represents the frequency of integers greater than or equal to the left endpoint of the interval but less than the right endpoint. For example, in the interval 10–20, the integers are greater than or equal to but less than .
Let be the mean of data set X and let be the mean of data set Y. Which choice is the smallest possible value of ?
When a histogram gives only intervals, the exact mean is not fixed, but it has a minimum and maximum. To make a mean as small as possible, place every value at the lowest integer in its interval; to make it as large as possible, place every value at the highest integer in its interval (right endpoint minus 1). For an expression like , minimize it by minimizing and maximizing , then subtract using the common denominator (here, 24).
Hints
Use endpoints, not midpoints
A histogram interval (like 20–30) only tells you the values are from to . To make a mean as small as possible, use the smallest values allowed; to make it as large as possible, use the largest values allowed.
Minimize a difference by pushing in opposite directions
To make as small as possible, try to make small and large at the same time.
Turn each mean into a weighted average
Compute a possible total sum for each data set by multiplying each interval’s chosen value by its frequency, then divide by 24.
Desmos Guide
Enter the endpoint sums for X
In Desmos, enter the expression 2*10+5*20+9*30+8*40 to get the smallest possible total for X.
Enter the endpoint sums for Y
Enter the expression 4*29+6*39+10*49+4*59 to get the largest possible total for Y.
Compute the difference of means
Enter (710/24)-(1076/24) (or use the totals you computed in steps 1 and 2). Then convert the result to a fraction to match the answer choices.
Step-by-step Explanation
Decide how to minimize
Because gets smaller when decreases and/or increases, the smallest possible value occurs when:
- is as small as possible (choose the smallest integers in each interval of X)
- is as large as possible (choose the largest integers in each interval of Y)
Compute the smallest possible mean of X
From the histogram for X, the frequencies are:
- 10–20: 2 values
- 20–30: 5 values
- 30–40: 9 values
- 40–50: 8 values
To make the mean as small as possible, use the left endpoints: .
Minimum possible sum for X:
So the smallest possible mean of X is .
Compute the largest possible mean of Y
From the histogram for Y, the frequencies are:
- 20–30: 4 values
- 30–40: 6 values
- 40–50: 10 values
- 50–60: 4 values
To make the mean as large as possible, use the greatest integers in each interval: .
Maximum possible sum for Y:
So the largest possible mean of Y is .
Subtract to get the smallest possible value
Therefore, the smallest possible value of is .