Super-Hard SAT Data Analysis Question 4 | Free SAT Prep

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Question 4·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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The frequency table summarizes the number of training sessions attended by each employee in a department. The value of $k$ is a positive integer, and the mean number of sessions in the original data set is $4.1$ .

Number of sessions	Frequency
$1$	$3$
$2$	$5$
$4$	$k$
$7$	$4$
$9$	$2$

In a corrected version of the data set, each record of $1$ session is changed to $3$ sessions, and each record of $9$ sessions is changed to $5$ sessions. Which choice gives the median and mean of the corrected data set?

For a frequency-table question, first translate the table into a total number of values and a total sum. Use the given mean to find any unknown frequency before analyzing changes. When values are corrected, track the change in the total rather than rebuilding the entire sum. For the median, sort the updated values conceptually and use cumulative frequencies to identify the middle position or positions.

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Write the total and count

Use the frequencies to express the original total number of sessions and the original number of employees in terms of $k$ .

Track changes to the total

The corrections do not change the number of data values. Determine how much each type of correction changes the total.

Use median positions

After finding the total number of values, identify the two middle positions in the corrected data set and use cumulative frequencies to locate them.

Desmos Guide

Verify the unknown frequency

Algebra is faster for this question. To verify $k$ in Desmos, graph $y_1=\frac{59+4x}{14+x}$ and $y_2=4.1$ . The intersection has an $x$ -coordinate of $16$ .

Verify the corrected mean

In a Desmos expression line, enter $(59+4(16)+3(3-1)+2(5-9))/(14+16)$ . This checks the corrected mean. Determine the median separately by using cumulative frequencies to locate the two middle positions.

Step-by-step Explanation

Use the original mean to find $k$

The original data set has $14+k$ values. Its total number of sessions is

\begin{aligned} 3(1)+5(2)+k(4)+4(7)+2(9) &= 59+4k. \end{aligned}

Use the given mean:

\begin{aligned} \frac{59+4k}{14+k} &= 4.1 \\ 10(59+4k) &= 41(14+k) \\ 590+40k &= 574+41k \\ k &= 16. \end{aligned}

Find the corrected mean

There are $14+16=30$ values, and the original total is $59+4(16)=123$ .

Changing three $1$ s to $3$ s increases the total by $3(2)=6$ . Changing two $9$ s to $5$ s decreases the total by $2(4)=8$ . Thus, the corrected total is $123+6-8=121$ .

Therefore, the corrected mean is

\frac{121}{30}.

Locate the middle values

The corrected data values, in ascending order by frequency, are $2$ occurring $5$ times, $3$ occurring $3$ times, $4$ occurring $16$ times, $5$ occurring $2$ times, and $7$ occurring $4$ times.

Because there are $30$ values, the median is the average of the $15$ th and $16$ th values. Both positions are among the sixteen $4$ s, so the median is $4$ .

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Step-by-step Explanation

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Step-by-step Explanation