Super-Hard SAT Data Analysis Question 13 | Free SAT Prep

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Question 13·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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The table summarizes the practice times, in minutes, for a data set of students. Each student's practice time is one of the values shown.

Practice time, in minutes	Number of students
$10$	$3$
$20$	$a$
$30$	$8$
$40$	$b$
$50$	$4$

The mean practice time is $29$ minutes, and the median practice time is $25$ minutes. The values of $a$ and $b$ are positive integers.

How many students had a practice time of $40$ minutes?

Your answer

When a frequency table gives both a mean and a median, use the median first if it identifies the middle position or positions. That often creates a relationship between unknown frequencies. Then use the mean as a weighted average, with each data value multiplied by its frequency, to solve for the remaining unknown.

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Interpret the median

Because $25$ is halfway between $20$ and $30$ , think about which two values must occupy the middle positions of the ordered data set.

Count one half of the data

Use the fact that the number of values at or below $20$ must equal half the total number of values.

Use a weighted mean

For the mean, multiply each practice time by its frequency, add those products, and divide by the total frequency.

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Set up the relationship from the median

Algebra is faster for this question. First use the median reasoning to obtain $b=a-9$ .

Graph the resulting equation

Let $x$ represent $a$ . Graph the two lines

y=60x+110

and

y=58x+174

Their intersection gives the value of $a$ .

Find the frequency for $40$ minutes

Subtract $9$ from the intersection's $x$ -coordinate to find $b$ , the number of students with a practice time of $40$ minutes.

Step-by-step Explanation

Use the median

Since $25$ is not one of the practice times, the two middle values must be $20$ and $30$ . Therefore, exactly half of the data values are at or below $20$ .

There are $3+a$ values at or below $20$ , and there are $15+a+b$ values total. Thus,

3+a=\frac{15+a+b}{2}

Solving gives $b=a-9$ .

Write the mean equation

The total of all practice times is

10(3)+20a+30(8)+40b+50(4)=470+20a+40b

Because the mean is $29$ ,

\frac{470+20a+40b}{15+a+b}=29

Substitute and solve

Substitute $b=a-9$ into the mean equation:

\frac{110+60a}{6+2a}=29

So $110+60a=174+58a$ , giving $a=32$ . Therefore,

b=32-9=23

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Step-by-step Explanation

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Step-by-step Explanation