Describing and comparing distributions practice

Check the count first. $5 + 8 + 6 + 4 + 0 + 0 + 0 + 1 = 24$ students, which matches the homeroom size, so no responses are missing.

Shape. The tallest stack sits at 1 pet, and from there the stacks fall away steadily (8 students, then 6, then 4) before three empty values and a single dot far out at 7. One peak with a long thin tail toward the large values makes the distribution unimodal and skewed to the right.

Center. With $n = 24$ the median is the average of the 12th and 13th values in order. The 5 students with 0 pets fill positions 1 through 5 and the 8 students with 1 pet fill positions 6 through 13, so both the 12th and the 13th values are 1, and the median is 1 pet. The mean is $\bar{x} = \frac{5(0) + 8(1) + 6(2) + 4(3) + 1(7)}{24} = \frac{39}{24} = 1.625$ pets.

Spread. The counts run from 0 pets to 7 pets, a range of 7 pets. For the middle half, positions 1 through 12 are 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, so $Q_1 = 1$, and positions 13 through 24 are 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 7, so $Q_3 = 2$. Then $IQR = 2 - 1 = 1$ pet.

Unusual features. No student reported 4, 5, or 6 pets, so there is a gap of three whole values before the largest observation. Testing it, $1.5 \times IQR = 1.5 \times 1 = 1.5$, so the upper fence is $Q_3 + 1.5 = 2 + 1.5 = 3.5$ pets, and the value 7 sits above it and is an outlier.

Put the four pieces together in context. The number of pets per household in this homeroom is unimodal and skewed to the right, centered at a median of 1 pet, with the middle 50% of students spanning just 1 pet ($IQR = 1$) out of a full range of 7 pets. One student reported 7 pets, an outlier above the 3.5-pet fence and separated from everyone else by a gap at 4, 5, and 6 pets. Because the distribution is skewed and carries an outlier, the median of 1 pet describes a typical household better than the mean of 1.625 pets.

Sort the rents: 850, 875, 880, 890, 900, 910, 925, 950, 2400. There are $n = 9$ listings.

Median. With 9 values the median is the 5th, so the median rent is 900 dollars.

Mean. The sum is $850 + 875 + 880 + 890 + 900 + 910 + 925 + 950 + 2400 = 9580$, so $\bar{x} = \frac{9580}{9} = 1064.44$ dollars. That is the number the blog quoted.

Quartiles and IQR. The count is odd, so leave the median out of both halves. The lower half is 850, 875, 880, 890, giving $Q_1 = \frac{875 + 880}{2} = 877.5$, and the upper half is 910, 925, 950, 2400, giving $Q_3 = \frac{925 + 950}{2} = 937.5$. Then $IQR = 937.5 - 877.5 = 60$ dollars.

Standard deviation. Working from the exact mean $\frac{9580}{9}$, the squared deviations sum to $2013472.22$, so $s = \sqrt{\frac{2013472.22}{9 - 1}} = \sqrt{251684.03} = 501.68$ dollars.

Unusual features. $1.5 \times IQR = 1.5 \times 60 = 90$, so the upper fence is $937.5 + 90 = 1027.5$ dollars. The 2400-dollar listing sits far above that fence and is an outlier, while the other eight rents all fall within 100 dollars of one another.

Decide, and say why. One extreme listing makes the distribution strongly skewed to the right. It has pulled the mean up to 1064.44 dollars, which is higher than 8 of the 9 rents on the list, so the blog's average describes no apartment a student could actually rent. It has also inflated $s$ to 501.68 dollars, more than five times the whole 100-dollar spread of the other eight. Report the resistant pair instead: a typical one-bedroom near this campus rents for a median of 900 dollars, and the middle half of the listings span only 60 dollars ($IQR = 60$).

(a) Shape. Read the heights across the classes: 3, 15, 10, 4, 9, 14, 5. They climb to a peak of 15 at 10 to 20 minutes, fall to a low of 4 at 30 to 40 minutes, climb again to a second peak of 14 at 50 to 60 minutes, then fall away. Two clear peaks separated by a dip make the distribution bimodal. Side to side it is roughly balanced rather than skewed: the three classes below 30 minutes hold $3 + 15 + 10 = 28$ workouts and the three above 40 minutes hold $9 + 14 + 5 = 28$.

(b) Median class. The total is $3 + 15 + 10 + 4 + 9 + 14 + 5 = 60$ workouts, so the median is the average of the 30th and 31st values in order. Running the cumulative counts gives 3 through the first class, 18 through the second, 28 through the third, and 32 through the fourth. Positions 29 through 32 therefore all sit in the 30 to 40 minute class, so both the 30th and the 31st values do, and the median lies in the 30 to 40 minute class.

(c) Percent under 20 minutes. The first two classes hold $3 + 15 = 18$ workouts, and $\frac{18}{60} = 0.30$, so 30% of the workouts were shorter than 20 minutes.

(d) Why one center misleads. The median lands in the 30 to 40 minute class, the shortest bar between the two peaks, holding only 4 of the 60 workouts. The center of this distribution falls in the valley between the two peaks, where almost nothing actually happened, so calling roughly 35 minutes a typical workout describes a length this user almost never trained for.

(d) Say what to do instead, in context. Two peaks usually mean two kinds of thing have been mixed into one graph, here plausibly short weekday sessions clustered around 10 to 20 minutes and long weekend sessions clustered around 50 to 60 minutes. The honest summary names both clusters and their centers rather than averaging across them.

(a) Shape from the mean against the median. Farm A: the mean of 142 g sits just 1 g above the median of 141 g, which is consistent with a roughly symmetric distribution. Farm B: the mean of 138 g sits 8 g above the median of 130 g, and a mean pulled well above the median is the signature of a right skew, almost certainly caused by a few unusually heavy tomatoes.

(b) Variability, three ways. Standard deviations: 11 g for Farm A against 24 g for Farm B. Interquartile ranges: $149 - 135 = 14$ g for A against $145 - 124 = 21$ g for B. Ranges: $168 - 118 = 50$ g for A against $226 - 110 = 116$ g for B. All three measures agree, so Farm B's weights are clearly the more variable.

(c) Outlier check, Farm A. $IQR = 14$ g, so $1.5 \times IQR = 21$ g. The fences are $135 - 21 = 114$ g and $149 + 21 = 170$ g. The minimum of 118 g and the maximum of 168 g both fall inside those fences, so the rule flags no outliers at Farm A.

(c) Outlier check, Farm B. $IQR = 21$ g, so $1.5 \times IQR = 31.5$ g. The fences are $124 - 31.5 = 92.5$ g and $145 + 31.5 = 176.5$ g. The minimum of 110 g is inside, but the maximum of 226 g is far above 176.5 g, so Farm B has at least one high outlier. A five-number summary cannot say how many tomatoes clear the fence, only that the heaviest one does.

(d) Comparison in context. The tomatoes sampled from Farm A are typically heavier than those sampled from Farm B, with a median weight of 141 g against 130 g, a difference of 11 g. The Farm A sample is also far more consistent, with an IQR of 14 g against Farm B's 21 g and a standard deviation of 11 g against 24 g. Farm A's weights look roughly symmetric with no outliers by the 1.5 IQR rule, while Farm B's are skewed to the right with at least one tomato above its 176.5 g fence, running as heavy as 226 g.

(a) Sort the first seven counts: 3, 4, 5, 5, 6, 6, 7. The sum is 36, so $\bar{x} = \frac{36}{7} = 5.14$ deer. With 7 values the median is the 4th, so the median is 5 deer.

(a) Quartiles and range. Leaving the median out, the lower half is 3, 4, 5 and the upper half is 6, 6, 7, so $Q_1 = 4$ and $Q_3 = 6$, giving $IQR = 6 - 4 = 2$ deer. The range is $7 - 3 = 4$ deer.

(a) Standard deviation. The squared deviations from $\frac{36}{7}$ sum to $10.857$, so $s = \sqrt{\frac{10.857}{7 - 1}} = \sqrt{1.810} = 1.35$ deer.

(b) Add the eighth night. Sorted, the counts are 3, 4, 5, 5, 6, 6, 7, 26. The sum is 62, so $\bar{x} = \frac{62}{8} = 7.75$ deer, and with an even count the median is the average of the 4th and 5th values, $\frac{5 + 6}{2} = 5.5$ deer.

(b) Quartiles, range, and standard deviation. The median falls between two values, so the lower half is 3, 4, 5, 5 and the upper half is 6, 6, 7, 26. Then $Q_1 = \frac{4 + 5}{2} = 4.5$ and $Q_3 = \frac{6 + 7}{2} = 6.5$, so $IQR = 6.5 - 4.5 = 2$ deer, exactly what it was before. The range jumps to $26 - 3 = 23$ deer. The squared deviations from 7.75 sum to $391.5$, so $s = \sqrt{\frac{391.5}{8 - 1}} = \sqrt{55.93} = 7.48$ deer.

(c) Line the two sets up. The mean climbed from 5.14 to 7.75 deer, a jump of 2.61 deer or about 51%. The standard deviation climbed from 1.35 to 7.48 deer, about 5.6 times as large. The range went from 4 to 23 deer. Against that, the median moved only from 5 to 5.5 deer and the IQR did not move at all.

(c) What it shows, in context. The mean and the standard deviation use every value, and the range uses the two extremes, so a single 26-deer night drags all three. The median and the IQR depend on position rather than size, so they absorbed the herd almost without flinching. Checking the rule on the eight nights, $1.5 \times IQR = 3$, so the upper fence is $6.5 + 3 = 9.5$ deer and the 26-deer night is an outlier. A description of what this camera sees on a typical night should therefore report the median of 5.5 deer with $IQR = 2$ deer, and mention the 26-deer herd separately as an outlier rather than letting it set the average.

Read the times off the plot in order: 18, 19, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 31, 32, 34, 36, 37, 58. That is $n = 20$ runners, and the stemplot has already sorted them.

Shape. Eleven of the twenty leaves sit on the stem of 2, the count thins to six leaves on the stem of 3, the stem of 4 carries no leaves at all, and then one leaf appears on the stem of 5. The distribution is unimodal with its peak in the 20s and skewed to the right, with a gap across the entire 40s.

Center. With 20 values the median is the average of the 10th and 11th times, $\frac{26 + 27}{2} = 26.5$ minutes. The mean is $\frac{565}{20} = 28.25$ minutes. The mean sitting 1.75 minutes above the median is a second signal of the right skew.

Spread. The range is $58 - 18 = 40$ minutes. For the quartiles, the lower 10 times are 18, 19, 20, 21, 22, 23, 24, 25, 25, 26, so $Q_1 = \frac{22 + 23}{2} = 22.5$ minutes, and the upper 10 are 27, 28, 29, 30, 31, 32, 34, 36, 37, 58, so $Q_3 = \frac{31 + 32}{2} = 31.5$ minutes. Then $IQR = 31.5 - 22.5 = 9$ minutes.

Outlier check. $1.5 \times IQR = 1.5 \times 9 = 13.5$ minutes, so the fences are $22.5 - 13.5 = 9$ minutes and $31.5 + 13.5 = 45$ minutes. The fastest time of 18 minutes is comfortably above the lower fence, but the slowest time of 58 minutes is well beyond the upper fence of 45 minutes, so it is an outlier.

Describe it in context. The finishing times for these 20 runners are unimodal and skewed to the right, centered at a median of 26.5 minutes, with the middle half of the field spanning 9 minutes ($IQR = 9$) inside a full range of 40 minutes. One runner finished in 58 minutes, an outlier above the 45-minute fence and cut off from the rest of the field by an empty stem across the 40s. Because of that outlier, the median and the IQR describe this race better than the mean of 28.25 minutes and the standard deviation would.

Sort each group. Standard: 6, 7, 7, 7, 8, 8, 9, 10, 12. Enriched: 5, 5, 6, 6, 6, 7, 7, 8, 14. Each group has $n = 9$ seeds.

Centers. With 9 values the median is the 5th. Standard has a median of 8 days and enriched has a median of 6 days. The means are $\frac{74}{9} = 8.22$ days for standard and $\frac{64}{9} = 7.11$ days for enriched.

Quartiles and spread, standard mix. Leaving the median out, the lower half is 6, 7, 7, 7 so $Q_1 = \frac{7 + 7}{2} = 7$, and the upper half is 8, 9, 10, 12 so $Q_3 = \frac{9 + 10}{2} = 9.5$. Then $IQR = 9.5 - 7 = 2.5$ days and the range is $12 - 6 = 6$ days.

Quartiles and spread, enriched mix. The lower half is 5, 5, 6, 6 so $Q_1 = \frac{5 + 6}{2} = 5.5$, and the upper half is 7, 7, 8, 14 so $Q_3 = \frac{7 + 8}{2} = 7.5$. Then $IQR = 7.5 - 5.5 = 2$ days and the range is $14 - 5 = 9$ days.

Unusual features. Standard: $1.5 \times 2.5 = 3.75$, so the upper fence is $9.5 + 3.75 = 13.25$ days, and the slowest seed at 12 days falls below it, so this group has no outliers. Enriched: $1.5 \times 2 = 3$, so the upper fence is $7.5 + 3 = 10.5$ days, and the seed that took 14 days is above it, so the enriched group has one high outlier.

Shape. In both groups the mean sits above the median (8.22 against 8, and 7.11 against 6) and the largest value stretches much farther above the median than the smallest falls below it, so both distributions are skewed to the right, the enriched group more strongly because of its outlier.

Resolve the conflicting spread measures before writing. The enriched group has the smaller IQR (2 days against 2.5 days) but the larger range (9 days against 6 days). Both are correct, and they are measuring different things: across the middle 50% of seeds the enriched mix was the more consistent, while its wider range comes entirely from the single 14-day seed. Name the measure whenever you make a spread claim.

Write the comparison in context. Seeds in the enriched mix germinated faster than seeds in the standard mix: the enriched median of 6 days is 2 days shorter than the standard median of 8 days. Both distributions are skewed to the right. Across the middle half of each group the enriched seeds were slightly more consistent, with an IQR of 2 days against the standard mix's 2.5 days, but the enriched group also holds one outlier at 14 days, above its 10.5-day fence, which stretches its range to 9 days against the standard mix's 6 days. The standard mix has no outliers by the 1.5 IQR rule.

(a) Compare the medians, which is the measure of center a boxplot displays. Store B's median wage of 21.00 dollars an hour is higher than Store A's median of 18.00, so Store B typically pays $21.00 - 18.00 = 3.00$ dollars an hour more.

(b) Compare two measures of spread. Interquartile ranges: Store A has $20.00 - 16.50 = 3.50$ dollars and Store B has $23.50 - 17.00 = 6.50$ dollars. Ranges: Store A has $24.00 - 15.00 = 9.00$ dollars and Store B has $26.00 - 14.00 = 12.00$ dollars. Both measures point the same way, so Store B's wages are the more variable.

(c) The two values named, 16.50 and 20.00, are exactly Store A's $Q_1$ and $Q_3$, and the box holds about the middle 50% of the data. So about 50% of Store A's 32 employees, roughly $0.50 \times 32 = 16$ employees, earn between 16.50 and 20.00 dollars an hour.

(d) Evaluate the claim against the plot. Store B's minimum wage is 14.00 dollars an hour, which is below Store A's minimum of 15.00. So Store B's lowest-paid employee earns less than every single Store A employee, not more than a quarter of them, and the claim is false. One number settles it: Store B's left whisker starts to the left of Store A's.

(e) Bimodality. No, these boxplots cannot settle it. A boxplot is built from exactly five numbers, so it shows center, spread, and the rough direction of skew while hiding everything about the interior of the data. Peaks, gaps, and clusters are all invisible, and two very different distributions can produce identical boxplots. Deciding whether Store B's wages are bimodal, which would be plausible if the store pays one rate to cashiers and a higher rate to supervisors, needs a dotplot, a stemplot, or a histogram of the individual wages.

Note what else the display withholds. Neither mean can be recovered from a five-number summary, so nothing here supports a claim about average pay. And because the whiskers were drawn to the extremes, no point on either plot has been flagged as an outlier. Applying the rule by hand: Store A's fences are $16.50 - 1.5(3.50) = 11.25$ and $20.00 + 1.5(3.50) = 25.25$ dollars, and Store B's are $17.00 - 1.5(6.50) = 7.25$ and $23.50 + 1.5(6.50) = 33.25$ dollars. Every reported value sits inside its own store's fences, so neither store shows an outlier by the 1.5 IQR rule.