Exploring One-Variable Data: AP Statistics Math Worksheet Answer Key – PDF Download

AP Statistics One Worksheet Answer Key: problems 1–5

Problem 1: Classify variables

An area code is a label, so arithmetic on it is meaningless.

1. Eye color: categorical.
2. Minutes of homework: quantitative.
3. Area code: categorical, because it is a label even though it looks like a number.
4. Number of pets: quantitative.

Problem 2: Relative frequencies

Divide each count by 50 and convert to a percent.

• Dog: \(\frac{21}{50}=42\%\)
• Cat: \(\frac{15}{50}=30\%\)
• Fish: \(\frac{6}{50}=12\%\)
• Bird: \(\frac{3}{50}=6\%\)
• None: \(\frac{5}{50}=10\%\)

Check: \(42+30+12+6+10=100\). Dog 42%, cat 30%, fish 12%, bird 6%, none 10%.

Problem 3: Reading a bar chart

1. \(\frac{126}{300}=0.42\), so 42% ride the bus.
2. Walk plus bike is \(54+30=84\), and \(\frac{84}{300}=0.28\). 28% neither ride the bus nor take a car.

Problem 4: Mean and median

Mean: \(\,\overline{x}=\frac{12+18+9+22+15+18+25}{7}=\frac{119}{7}=17\).

Ordered: 9, 12, 15, 18, 18, 22, 25. The middle (fourth) value is 18.

Mean 17 points, median 18 points.

Problem 5: Effect of an outlier

Before: \(\,\overline{x}=\frac{86}{5}=17.2\) and the median is 17.

After: \(\,\overline{x}=\frac{166}{6}\approx 27.67\). The ordered values are 14, 16, 17, 19, 20, 80, so the median is \(\frac{17+19}{2}=18\).

The mean rose by about 10.5 minutes while the median rose by 1. The median is resistant to the outlier; the mean is not.

AP Statistics One Worksheet Answer Key: problems 6–10

Problem 6: Range and IQR

The median is the sixth value, 13. Leave it out. The lower five values 4, 7, 9, 10, 12 have median \(Q_1=9\). The upper five 15, 18, 19, 22, 30 have median \(Q_3=19\).

\(\text{IQR}=19-9=10\) and \(\text{range}=30-4=26\). Range 26, IQR 10.

Problem 7: Five-number summary

The 12 values are already ordered. The median is the mean of the sixth and seventh values: \(\frac{22+25}{2}=23.5\).

Lower half 8, 12, 15, 15, 18, 22 has median \(\frac{15+15}{2}=15\). Upper half 25, 27, 31, 34, 40, 52 has median \(\frac{31+34}{2}=32.5\).

Five-number summary: 8, 15, 23.5, 32.5, 52.

\(\text{IQR}=32.5-15=17.5\), so the fences are \(15-1.5(17.5)=-11.25\) and \(32.5+1.5(17.5)=58.75\). All values lie inside, so there are no outliers.

Problem 8: Sample standard deviation

\(\,\overline{x}=\frac{55}{5}=11\). The deviations are \(-5, -2, -1, 2, 6\) and their squares are \(25, 4, 1, 4, 36\), with sum 70.

\(s=\sqrt{\frac{70}{5-1}}=\sqrt{17.5}\approx 4.18\). s is about 4.18.

Problem 9: Shape and median from a histogram

1. The graph has one peak at 40 to 60 minutes and a longer tail toward the large values: unimodal and skewed right.
2. The 20th and 21st ordered values locate the median. The cumulative counts are 4, 15, 28, so both lie in the 40 to 60 class: 40 to 60 minutes.
3. Skewed right means the long tail pulls the mean toward large values, so the mean is likely greater than the median.

Problem 10: Counting from a histogram

1. \(2+5=7\), so 7 students scored below 70.
2. \(9+4=13\) students scored 80 or higher, and \(\frac{13}{30}\approx 0.433\). About 43.3%.
3. The cumulative counts are 2, 7, 17. The 15th and 16th values lie in the third class: 70 to 80.

AP Statistics One Worksheet Answer Key: problems 11–15

Problem 11: Test for outliers with IQR

Lower five values 22, 24, 25, 27, 28 give \(Q_1=25\). Upper five 29, 31, 33, 34, 62 give \(Q_3=33\).

\(\text{IQR}=33-25=8\), and \(1.5\times 8=12\).

Upper fence: \(33+12=45\). Lower fence: \(25-12=13\).

Since \(62 > 45\), 62 is an outlier. No other value is outside 13 to 45.

Problem 12: Reading a boxplot

1. Median 7, \(Q_1=5\), \(Q_3=9\) (thousand steps).
2. \(\text{IQR}=9-5=4\) and the upper fence is \(9+1.5(4)=15\). Since \(17 > 15\), 17 is an outlier. The upper whisker ends at 13, the largest value inside the fence.
3. The box holds the middle half of the data: about 50%.

Problem 13: Compute a z-score

1. \(z=\frac{84-72}{6}=2\). z = 2: the score is 2 standard deviations above the mean.
2. \(x=72+(-1.5)(6)=63\). The score was 63.

Problem 14: Compare two z-scores

Jamal: \(z=\frac{78-70}{5}=1.6\). Sofia: \(z=\frac{85-75}{8}=1.25\).

Jamal did better relative to his class, because his score is 1.6 standard deviations above the mean and Sofia’s is 1.25.

Problem 15: Percentiles

1. \(\frac{21}{25}=0.84\), so Ethan is at the 84th percentile.
2. \(0.60\times 40=24\). 24 students scored below Maya.

AP Statistics One Worksheet Answer Key: problems 16–20

Problem 16: Empirical rule percentages

1. \(12=14-1(2)\) and \(18=14+2(2)\). The area from \(-1\) to \(2\) standard deviations is \(34\%+34\%+13.5\%=81.5\%\). About 81.5%.
2. Beyond 2 standard deviations on the high side lies \(\frac{100\%-95\%}{2}=2.5\%\). About 2.5%.

Problem 17: Normal probability above a value

\(z=\frac{15-12.0}{2.5}=1.2\).

The area to the left of \(z=1.2\) is 0.8849, so the area to the right is \(1-0.8849=0.1151\).

About 11.5% of deliveries take more than 15 minutes.

Problem 18: Value at a percentile

For a left area of 0.90, the z-score is \(z\approx 1.28\) (invNorm(0.90) = 1.2816).

\(x=500+1.2816(100)\approx 628.2\). The 90th percentile is about 628.

Problem 19: Compare two distributions

Center: the median for Group 2 (23 minutes) is 6 minutes higher than for Group 1 (17 minutes), so Group 2 typically took longer.

Spread: the IQR is 12 minutes for Group 2 and 8 minutes for Group 1, and the ranges are 29 and 22 minutes, so Group 2 times vary more.

Shape: Group 1 has a longer upper whisker (the median is nearer Q1), which suggests a distribution skewed right; Group 2 is close to symmetric. Neither group has an outlier (upper fences 34 and 47).

Problem 20: Two outlier rules

1. \(Q_1=\frac{4+5}{2}=4.5\), \(Q_3=\frac{7+8}{2}=7.5\), \(\text{IQR}=3\). The upper fence is \(7.5+1.5(3)=12\) and the lower fence is \(4.5-4.5=0\). Only 20 is an outlier.
2. \(\,\overline{x}=\frac{84}{12}=7\) and \(s\approx 4.45\). The limits are \(7.5\pm 2(4.45)\), that is \(-1.90\) to \(15.90\). Only 20 is outside, so it is the only outlier under this rule too.

AP Statistics One Worksheet Answer Key: problems 21–22

Problem 21: Changing units

Multiplying by 1.8 multiplies the mean, median, standard deviation and IQR by 1.8; adding 32 shifts only the measures of center.

• Mean: \(1.8(20)+32=68\)
• Median: \(1.8(19)+32=66.2\)
• Standard deviation: \(1.8(4)=7.2\)
• IQR: \(1.8(6)=10.8\)

Mean 68 °F, median 66.2 °F, standard deviation 7.2 °F, IQR 10.8 °F.

Problem 22: Salaries at a bakery

1. \(\,\overline{x}=\frac{468{,}000}{9}=\$52{,}000\). The ordered middle (fifth) value gives a median of \(\$36{,}500\).
2. \(Q_1=\frac{33{,}500+34{,}000}{2}=\$33{,}750\) and \(Q_3=\frac{39{,}000+41{,}000}{2}=\$40{,}000\), so \(\text{IQR}=\$6{,}250\). The upper fence is \(40{,}000+1.5(6{,}250)=\$49{,}375\). Since \(180{,}000 > 49{,}375\), the owner’s salary is an outlier.
3. The median, $36,500. The mean of $52,000 is higher than 8 of the 9 salaries because the owner’s pay pulls it up, while the median is resistant.

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