AP Statistics Mean, Median, Standard Deviation and Outliers Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Mean, Median, Standard Deviation and Outliers – Maths PDF

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This summary statistics answers for AP Statistics covers mean, median, standard deviation and outliers (Common Core AP STAT 1.6, AP STAT 1.7, HSS-ID.A.2, HSS-ID.A.3), with a free PDF to print.

Each solution shows the sorted data, every deviation or fence, and the final value in context. Use the answers to check intermediate steps, not only the last line.

AP Statistics · Common Core State Standards: AP STAT 1.6, AP STAT 1.7, HSS-ID.A.2, HSS-ID.A.3

Summary statistics answers: exercises 1 to 10

Exercise 1 – Compute a mean

$\overline{x}=\frac{12+15+9+14+20}{5}=\frac{70}{5}=14.0$.

14.0 points.

Exercise 2 – Median of an odd list

Order: 17, 19, 22, 23, 26, 28, 31.

The 4th value of 7 is the median.

23 minutes.

Exercise 3 – Median of an even list

Order: 5, 6, 8, 9, 12, 14.

The middle values are 8 and 9, so $\frac{8+9}{2}=8.5$.

8.5.

Exercise 4 – Find the missing value

Total needed: $5\times 84=420$. Known total: $90+78+85+81=334$.

Fifth score: $420-334=86$.

86.

Exercise 5 – Range and IQR

Median is 10. Lower half: 4, 6, 7, 9 so $Q_1=6.5$. Upper half: 12, 13, 15, 18 so $Q_3=14$.

Range: $18-4=14$. IQR: $14-6.5=7.5$.

Range 14, Q1 6.5, Q3 14, IQR 7.5.

Exercise 6 – Standard deviation by hand

Mean: 7. Deviations: -4, -2, 0, 2, 4. Squares: 16, 4, 0, 4, 16, sum 40.

$s=\sqrt{\frac{40}{4}}=\sqrt{10}\approx 3.16$.

3.16.

Exercise 7 – Effect of an outlier

Before the error, mean and median agree.

Before: mean $\frac{72}{6}=12.00$, median 12. After: mean $\frac{198}{6}=33.00$, median 12.

The mean jumps from 12.00 to 33.00 but the median stays at 12, so the median is resistant.

Exercise 8 – Mean versus median

The tail points right, so the mean is greater than the median.

$\overline{x}=\frac{54}{12}=4.5$.

Greater: the mean is 4.5.

Exercise 9 – 1.5 IQR rule

Fences for the data set.

$Q_1=7.5$, $Q_3=14$, $IQR=6.5$.

Fences: $7.5-1.5(6.5)=-2.25$ and $14+1.5(6.5)=23.75$.

Only 40 lies beyond a fence, so 40 is the lone outlier.

Exercise 10 – Resistant statistics

A resistant statistic changes little when extreme values are added or changed.

Resistant: median and IQR. Not resistant: mean, range, and standard deviation.

Exercises 11 and up

Exercise 11 – Add a constant

Adding 5 shifts the center and quartiles by 5 and leaves the spread unchanged.

Mean 77, median 79, standard deviation 8, IQR 11.

Exercise 12 – Multiply by a constant

Mean: $1.8(20)+32=68$. The addition of 32 does not change the spread, so $s=1.8\times 3=5.4$.

Mean 68°F; standard deviation 5.4°F.

Exercise 13 – Compare spreads

Data set A.

A: squares of deviations $4+1+1+0+0+0+1+1+4=12$, $s=\sqrt{12/8}\approx 1.22$.

B: sum of squares $16+9+4+1+0+1+4+9+16=60$, $s=\sqrt{60/8}\approx 2.74$.

B has the larger standard deviation (2.74 compared with 1.22) because its values are more spread out.

Exercise 14 – Mean from a frequency table

Pets owned by 25 students.

Total students: $5+8+6+6=25$. Total pets: $0(5)+1(8)+2(6)+3(6)=38$.

Mean: $\frac{38}{25}=1.52$. The 13th value of 25 is in the 1-pet group (5 + 8 = 13), so the median is 1.

Mean 1.52 pets; median 1 pet.

Exercise 15 – Which measure to report?

Sum: 534; mean $\frac{534}{7}\approx 76.3$. Median: the 4th value, 48.

Mean about $76,300; median $48,000. The median is better because the 250 outlier inflates the mean.

Exercise 16 – Combine two means

Total points: $20(80)+30(90)=1600+2700=4300$.

$\frac{4300}{50}=86$.

86.

Exercise 17 – Required score

Total needed: $5\times 80=400$. Current total: $72+78+85+77=312$.

$400-312=88$.

At least 88.

Exercise 18 – Full summary

Median $\frac{17+18}{2}=17.5$. $Q_1=15$, $Q_3=21$, $IQR=6$.

Fences: $15-9=6$ and $21+9=30$.

Summary 14, 15, 17.5, 21, 41; the value 41 is an outlier.

Exercise 19 – Describe with statistics

Mean: $\frac{579}{9}=64.3$, greater than the median 61.

The mean is pulled above the median by the 95-pound dog, so the data are skewed right and the median is the better summary.

Exercise 20 – Adding a value

New mean: $\frac{120+50}{6}=\frac{170}{6}\approx 28.3$. New median: ordered 20, 22, 25, 25, 28, 50, so $\frac{25+25}{2}=25$.

The mean rises from 24 to about 28.3; the median stays 25. The mean changed more.

Key facts behind this summary statistics answers

Key facts

Mean = sum divided by n; median = middle of the ordered list.
Skewed right: mean > median; skewed left: mean < median; symmetric: about equal.
s = sqrt( sum of squared deviations / (n – 1) ); IQR = Q3 – Q1.
1.5 IQR rule: outliers lie below Q1 – 1.5 IQR or above Q3 + 1.5 IQR.
Adding c shifts center but not spread; multiplying by k scales both by |k|.

How to use this summary statistics answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the summary statistics answers as a PDF with the button below and keep it with your notes.

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