A data set contains 20 values. The number 5 appears 8 times, the number 9 appears 10 times, and the number appears 2 times. Which choice of results in the smallest standard deviation of the data set?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 55 | Free SAT Prep | Aniko

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Question 55·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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A data set contains 20 values. The number 5 appears 8 times, the number 9 appears 10 times, and the number $x$ appears 2 times.

Which choice of $x$ results in the smallest standard deviation of the data set?

When a data set includes an unknown value and you’re asked to minimize standard deviation, rewrite the spread using an expression you can optimize. A fast SAT approach is to use $\sum (v-\mu)^2=\sum v^2-\frac{(\sum v)^2}{n}$ to build a quadratic in $x$ and then use the vertex formula $x=-\frac{b}{2a}$ , avoiding long expansion of individual deviations.

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Think about what standard deviation measures

Standard deviation measures how far values are from the mean, with larger distances counting more because they are squared.

Use an efficient variance formula

Instead of expanding many $(v-\mu)^2$ terms, use $\sum (v-\mu)^2=\sum v^2-\frac{(\sum v)^2}{n}$ with $n=20$ .

Once you have a quadratic, use the vertex

After you rewrite the variance numerator as $ax^2+bx+c$ with $a>0$ , the minimum occurs at $x=-\frac{b}{2a}$ .

Desmos Guide

Enter the quadratic for the variance numerator

In Desmos, define

T(x)=(9/5)x^2-26x+165.

Graph and find the minimum

Graph $y=T(x)$ . Because this is a parabola opening upward, its lowest point is the minimum.

Read the $x$ -coordinate of the vertex

Click the vertex (lowest point) of the parabola. The $x$ -coordinate shown is the value of $x$ that minimizes the variance, and therefore minimizes the standard deviation.

Step-by-step Explanation

Write the sum and sum of squares

Use the variance identity (for a full data set of size $n$ ):

\sum (v-\mu)^2 = \sum v^2 - \frac{(\sum v)^2}{n}

Here $n=20$ .

Sum of the values:

\sum v = 8\cdot 5 + 10\cdot 9 + 2\cdot x = 130+2x

Sum of squares:

\sum v^2 = 8\cdot 5^2 + 10\cdot 9^2 + 2\cdot x^2 = 1010+2x^2

Express the (numerator of) variance as a quadratic in $x$

Let

T(x)=\sum v^2-\frac{(\sum v)^2}{20}.

Then

T(x)=\left(1010+2x^2\right)-\frac{(130+2x)^2}{20}.

Simplifying gives a quadratic:

T(x)=\frac{9}{5}x^2-26x+165.

Because standard deviation is $\sqrt{\text{variance}}$ and variance is $T(x)/20$ , the value of $x$ that minimizes standard deviation is the same value that minimizes $T(x)$ .

Minimize the quadratic

A quadratic $ax^2+bx+c$ with $a>0$ is minimized at its vertex, whose $x$ -coordinate is

x=-\frac{b}{2a}.

For $T(x)=\frac{9}{5}x^2-26x+165$ , this gives

x=-\frac{-26}{2\cdot \frac{9}{5}}=\frac{26}{\frac{18}{5}}.

Match to an answer choice

Simplify:

\frac{26}{\frac{18}{5}}=26\cdot \frac{5}{18}=\frac{130}{18}=\frac{65}{9}.

So the choice of $x$ that yields the smallest standard deviation is $\frac{65}{9}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation