Comparing two distributions practice problems

Sort each brand. Brand K: 22, 23, 24, 24, 25, 25, 26, 27, 28. Brand M: 15, 18, 21, 22, 24, 27, 29, 30, 33. Each brand has $n = 9$ batteries.

Medians. With 9 values the median is the 5th value in sorted order, so Brand K has a median of 25 hours and Brand M has a median of 24 hours.

Brand K quartiles. The count is odd, so leave the median out of both halves. The lower half 22, 23, 24, 24 gives $Q_1 = (23 + 24)/2 = 23.5$ hours, and the upper half 25, 26, 27, 28 gives $Q_3 = (26 + 27)/2 = 26.5$ hours. Then $IQR = 26.5 - 23.5 = 3$ hours.

Brand M quartiles. The lower half 15, 18, 21, 22 gives $Q_1 = (18 + 21)/2 = 19.5$ hours, and the upper half 27, 29, 30, 33 gives $Q_3 = (29 + 30)/2 = 29.5$ hours. Then $IQR = 29.5 - 19.5 = 10$ hours.

Ranges, as a second look at spread: Brand K spans $28 - 22 = 6$ hours and Brand M spans $33 - 15 = 18$ hours, three times as wide.

Write each sentence with a linking word, both numbers, and the units. Center: the median run time for Brand K, 25 hours, is only 1 hour longer than the median for Brand M, 24 hours, so a typical battery of either brand lasts about the same time. Variability: Brand K is far more consistent than Brand M, with an IQR of 3 hours against Brand M's 10 hours and a range of 6 hours against 18 hours.

Check the counts. Instructor P: $1 + 3 + 5 + 6 + 3 + 2 = 20$ students. Instructor R: $2 + 3 + 2 + 4 + 3 + 2 + 2 + 1 + 1 = 20$ students. The two groups are the same size, so the dot counts can be compared directly.

Instructor P, center and quartiles. With $n = 20$ the median is the average of the 10th and 11th values. Cumulative counts are 1 through 8 hours, 4 through 9 hours, 9 through 10 hours, and 15 through 11 hours, so positions 10 and 11 both land on 11 hours and the median is 11 hours. The lower half is positions 1 to 10, so $Q_1$ is the average of positions 5 and 6, both 10 hours, giving $Q_1 = 10$. The upper half is positions 11 to 20, so $Q_3$ is the average of positions 15 and 16, which are 11 and 12 hours, giving $Q_3 = 11.5$. Then $IQR = 11.5 - 10 = 1.5$ hours.

Instructor R, center and quartiles. Cumulative counts are 2 through 9 hours, 5 through 10, 7 through 11, 11 through 12, 14 through 13, 16 through 14, 18 through 15, 19 through 16, and 20 through 22, so positions 10 and 11 both land on 12 hours and the median is 12 hours. $Q_1$ is the average of positions 5 and 6, which are 10 and 11 hours, giving $Q_1 = 10.5$. $Q_3$ is the average of positions 15 and 16, both 14 hours, giving $Q_3 = 14$. Then $IQR = 14 - 10.5 = 3.5$ hours.

Ranges: Instructor P spans $13 - 8 = 5$ hours and Instructor R spans $22 - 9 = 13$ hours.

Unusual features. Instructor P: $1.5 \times 1.5 = 2.25$, so the fences are $10 - 2.25 = 7.75$ and $11.5 + 2.25 = 13.75$ hours, and the minimum of 8 and the maximum of 13 both sit inside, so there are no outliers. Instructor R: $1.5 \times 3.5 = 5.25$, so the fences are $10.5 - 5.25 = 5.25$ and $14 + 5.25 = 19.25$ hours, and the student who logged 22 hours is above 19.25, so that student is a high outlier. No student of Instructor R logged 17, 18, 19, 20, or 21 hours, so there is a gap of five whole values before that outlier.

Shape. Instructor P's counts run 1, 3, 5, 6, 3, 2 with a single peak at 11 hours and a fall on both sides, and the mean is $\frac{8(1) + 9(3) + 10(5) + 11(6) + 12(3) + 13(2)}{20} = \frac{213}{20} = 10.65$ hours, close to the median of 11, so the distribution is unimodal and roughly symmetric. Instructor R's counts peak at 12 hours and trail off to the right, and the mean is $\frac{253}{20} = 12.65$ hours, above the median of 12, so that distribution is unimodal and skewed to the right.

Write the comparison in context. Among these observed first-time passers, Instructor P's students logged fewer practice hours at the median: P's median of 11 hours is 1 hour below R's median of 12 hours. P's students were also far more consistent, with an IQR of 1.5 hours against R's 3.5 hours and a range of 5 hours against R's 13 hours. P's distribution is roughly symmetric while R's is skewed to the right. P's students show no outliers by the 1.5 IQR rule, while R's group has one student at 22 hours, above R's upper fence of 19.25 hours and separated from the rest by a gap running from 17 to 21 hours.

(a) Interquartile ranges. Shop V: $IQR = Q_3 - Q_1 = 7.5 - 4.5 = 3$ minutes. Shop W: $IQR = 13 - 7 = 6$ minutes. Both numbers in the response check out.

(a) Outlier check, Shop V. $1.5 \times IQR = 1.5 \times 3 = 4.5$ minutes, so the fences are $4.5 - 4.5 = 0$ and $7.5 + 4.5 = 12$ minutes. The minimum of 2 minutes and the maximum of 11 minutes both sit inside, so Shop V shows no outliers.

(a) Outlier check, Shop W. $1.5 \times IQR = 1.5 \times 6 = 9$ minutes, so the fences are $7 - 9 = -2$ and $13 + 9 = 22$ minutes. The longest wait of 26 minutes is above 22, so it is a high outlier and the response is right about it.

(b) The response reports each shop separately. Make the relationships explicit by saying that W has a median wait 3 minutes longer than V and an IQR twice as large. Contrast the suggested shapes and note that only W has a flagged high value. These connections make the comparison easier to follow.

(b) Use direct comparative language for center and spread, supported by the matching numerical summaries. Shape and unusual features can be compared in words. Whether a response receives particular credit depends on the question and its rubric.

(c) Rewrite it one element at a time. Center: waits at Shop W are typically 3 minutes longer than at Shop V, with medians of 9 minutes and 6 minutes. Variability: Shop W's waits are twice as variable across the middle half, with an IQR of 6 minutes against Shop V's 3 minutes, and a range of 23 minutes against Shop V's 9 minutes.

(c) Shape: Shop V's waits are roughly symmetric while Shop W's are skewed to the right. Unusual features: Shop W has a high outlier at 26 minutes, above its upper fence of 22 minutes, while Shop V has none, since its fences of 0 and 12 minutes contain every wait it recorded.

(a) A boxplot displays the median, not the mean, so compare medians. The Catacomb's median escape time of 47 minutes is 1 minute longer than The Vault's median of 46 minutes, so teams take about the same time in the two rooms.

(b) Interquartile ranges: The Vault has $52 - 41 = 11$ minutes and The Catacomb has $57 - 42 = 15$ minutes. Ranges: The Vault has $60 - 32 = 28$ minutes and The Catacomb has $74 - 34 = 40$ minutes. Both measures point the same way, so The Catacomb's times are the more variable.

(c) Shape from the box and the whiskers. The Vault: the median sits 5 minutes above $Q_1$ and 6 minutes below $Q_3$, and the whiskers run 9 minutes down and 8 minutes up, so the display is close to balanced and the times look roughly symmetric.

(c) The Catacomb: the median sits 5 minutes above $Q_1$ but 10 minutes below $Q_3$, and the upper whisker runs 17 minutes against the lower whisker's 8 minutes, so those times are skewed to the right.

(d) The Vault: $1.5 \times IQR = 1.5 \times 11 = 16.5$ minutes, so the fences are $41 - 16.5 = 24.5$ and $52 + 16.5 = 68.5$ minutes. The fastest team at 32 minutes and the slowest at 60 minutes both sit inside, so nothing is plotted separately.

(d) The Catacomb: $1.5 \times IQR = 1.5 \times 15 = 22.5$ minutes, so the fences are $42 - 22.5 = 19.5$ and $57 + 22.5 = 79.5$ minutes. The slowest team at 74 minutes is below 79.5, so the whisker reaches it and again nothing is plotted separately.

(e) No. A boxplot is built from five numbers and the mean is not one of them. The Catacomb's right skew makes it likely that its mean sits above its median, but that is an expectation rather than a reading, and no calculation from these five numbers recovers either mean.

(f) Comparison in context. Teams escape the two rooms in about the same typical time, with The Catacomb's median of 47 minutes only 1 minute above The Vault's 46 minutes. The Catacomb's times are the more variable, with an IQR of 15 minutes against The Vault's 11 minutes and a range of 40 minutes against 28 minutes. The Vault's times are roughly symmetric while The Catacomb's are skewed to the right, stretching to a slowest team of 74 minutes. Neither room has an escape time flagged as an outlier by the 1.5 IQR rule.

Sort each list. Restaurant: 62, 78, 88, 92, 96, 104, 118, 132, 145, 175, 210. Bookstore: 86, 88, 90, 91, 92, 92, 93, 94, 95, 96, 98. Each job has $n = 11$ shifts.

(a) Restaurant. The median is the 6th value, 104 dollars. Leaving it out, the lower half 62, 78, 88, 92, 96 gives $Q_1 = 88$ dollars and the upper half 118, 132, 145, 175, 210 gives $Q_3 = 145$ dollars, so $IQR = 145 - 88 = 57$ dollars and the range is $210 - 62 = 148$ dollars.

(a) Bookstore. The median is the 6th value, 92 dollars. The lower half 86, 88, 90, 91, 92 gives $Q_1 = 90$ dollars and the upper half 93, 94, 95, 96, 98 gives $Q_3 = 95$ dollars, so $IQR = 95 - 90 = 5$ dollars and the range is $98 - 86 = 12$ dollars.

(b) The restaurant has the higher typical night: its median of 104 dollars is 12 dollars above the bookstore's median of 92 dollars. The bookstore earnings were much less variable over these observed shifts: the restaurant's IQR of 57 dollars is more than 11 times the bookstore's 5 dollars, and its range of 148 dollars is more than 12 times the bookstore's 12 dollars.

(c) For a projection, assume these observed means are representative of future shifts and multiply each by 60. Actual future totals can differ. Restaurant: the 11 nights sum to $62 + 78 + 88 + 92 + 96 + 104 + 118 + 132 + 145 + 175 + 210 = 1300$ dollars, so $\bar{x} = 1300/11 = 118.18$ dollars a night, where $\bar{x}$ (x-bar) is the sample mean. Bookstore: the 11 nights sum to $86 + 88 + 90 + 91 + 92 + 92 + 93 + 94 + 95 + 96 + 98 = 1015$ dollars, so $\bar{x} = 1015/11 = 92.27$ dollars a night.

(c) The restaurant averages $\frac{1300 - 1015}{11} = \frac{285}{11} \approx 25.91$ dollars more per shift, so, if that average difference continues, the projected advantage across 60 shifts is $60 \times \frac{285}{11} = \frac{17100}{11} \approx 1554.55$ dollars more. The restaurant has the higher projected total under that assumption, not a guaranteed higher total.

(d) This question is about the worst nights rather than the typical one, so look at the low tail instead of the center. At the bookstore all 11 nights cleared 85 dollars, the lowest being 86. At the restaurant 2 of the 11 nights came in under 85 dollars (62 and 78), about 18% of shifts, and the whole bottom quarter of restaurant nights sits at or below $Q_1 = 88$ dollars, barely above the line. The bookstore has the more favorable observed low tail, but 11 successful nights do not guarantee that every future night clears 85 dollars. An absolute minimum would require a guaranteed-pay arrangement, not just these observations.

(e) In these samples, the restaurant has a higher mean but lower observed worst nights. Compare the actual low tail rather than inferring it from center and spread alone. A higher center and wider spread do not always imply lower minimum earnings. The choice depends on the objective, the relevance of these observations to future shifts, and any guaranteed-pay terms.

(a) Check the totals first: $30 + 75 + 105 + 75 + 15 = 300$ students at Central High and $12 + 21 + 15 + 9 + 3 = 60$ at Riverside. Central surveyed five times as many students. Raw counts describe the numbers of surveyed students in each interval, but relative frequencies are needed to compare the distribution of reading time without conflating it with group size.

(a) Different vertical scales can make raw bar heights misleading to compare. Read each axis, then divide counts by its group total to compare relative frequencies.

(b) Central High relative frequencies: $30/300 = 0.10$, $75/300 = 0.25$, $105/300 = 0.35$, $75/300 = 0.25$, and $15/300 = 0.05$. They sum to 1.00.

(b) Riverside relative frequencies: $12/60 = 0.20$, $21/60 = 0.35$, $15/60 = 0.25$, $9/60 = 0.15$, and $3/60 = 0.05$. These also sum to 1.00, and the lowest class now reverses: 20% of Riverside students read under 15 minutes against only 10% at Central High, even though Riverside's raw count in that class, 12 students, is smaller than Central High's 30.

(c) Central High median class. With 300 students the median is the average of the 150th and 151st values in order. Cumulative counts are 30, 105, 210, 285, and 300, so positions 106 through 210 all fall in the 30 to 45 minute class and both the 150th and the 151st values sit there. The median class is 30 to 45 minutes.

(c) Riverside median class. With 60 students the median is the average of the 30th and 31st values. Cumulative counts are 12, 33, 48, 57, and 60, so positions 13 through 33 fall in the 15 to 30 minute class and both the 30th and the 31st values sit there. The median class is 15 to 30 minutes.

Shape and the upper tail. Central High's relative frequencies run 0.10, 0.25, 0.35, 0.25, 0.05, rising to a single peak in the 30 to 45 minute class and falling away on both sides, so that distribution is unimodal and close to symmetric. Riverside's run 0.20, 0.35, 0.25, 0.15, 0.05, peaking one class lower and then falling steadily, so that distribution is unimodal and skewed to the right. At the top end, $(75 + 15)/300 = 90/300 = 0.30$ of Central High students read at least 45 minutes against $(9 + 3)/60 = 12/60 = 0.20$ at Riverside.

(d) Comparison in context. Central High students typically read longer than Riverside students: Central High's median falls in the 30 to 45 minute class while Riverside's falls one class lower, in the 15 to 30 minute class. Central High's distribution is close to symmetric about that peak while Riverside's is skewed to the right, with twice the share of students reading under 15 minutes (20% against 10%). A larger share of Central High students also reads at least 45 minutes a day, 30% against 20%. Calculate these shares from the counts and group totals; do not compare unadjusted bar heights across the differently scaled graphs.

(a) Read the leaves off each side of the stems. North: 52, 55, 56, 58, 60, 61, 63, 64, 66, 68, 70, 74. South: 12, 46, 48, 50, 51, 53, 54, 55, 57, 59, 62, 65. Each array has $n = 12$ days, and the stemplot has already put both in order.

(b) With 12 values the median is the average of the 6th and 7th. North: $(61 + 63)/2 = 124/2 = 62$ kWh. South: $(53 + 54)/2 = 107/2 = 53.5$ kWh. The North array produced 8.5 kWh more on a typical day.

(c) Quartiles, North. The lower half is 52, 55, 56, 58, 60, 61, so $Q_1 = (56 + 58)/2 = 57$ kWh. The upper half is 63, 64, 66, 68, 70, 74, so $Q_3 = (66 + 68)/2 = 67$ kWh. Then $IQR = 67 - 57 = 10$ kWh and the range is $74 - 52 = 22$ kWh.

(c) Quartiles, South. The lower half is 12, 46, 48, 50, 51, 53, so $Q_1 = (48 + 50)/2 = 49$ kWh. The upper half is 54, 55, 57, 59, 62, 65, so $Q_3 = (57 + 59)/2 = 58$ kWh. Then $IQR = 58 - 49 = 9$ kWh and the range is $65 - 12 = 53$ kWh.

(c) The two measures disagree: South has the smaller IQR (9 kWh against North's 10 kWh) but a range more than twice as wide (53 kWh against 22 kWh). Both are correct because they measure different things. Across its middle 50% of days South is slightly the steadier array, while its much wider range comes entirely from the single 12 kWh day. Name the measure of spread you are using whenever you make a variability claim.

(d) North: $1.5 \times 10 = 15$ kWh, so the fences are $57 - 15 = 42$ and $67 + 15 = 82$ kWh. The lowest day at 52 kWh and the highest at 74 kWh both sit inside, so North has no outliers.

(d) South: $1.5 \times 9 = 13.5$ kWh, so the fences are $49 - 13.5 = 35.5$ and $58 + 13.5 = 71.5$ kWh. The 12 kWh day is far below 35.5, so it is a low outlier, and the empty stems at 2 and 3 show a gap of more than 30 kWh separating it from South's next lowest day at 46 kWh.

(e) These 12 days show different output patterns: North has a median 8.5 kWh higher and a range of 22 kWh rather than South's 53 kWh. South has an unusually low recorded-output day at 12 kWh, below its 35.5 kWh fence; North has no flagged day. Those facts do not establish why the low output occurred or compare long-run reliability. Equipment operation, weather, shading, and measurement issues would require additional evidence.

Both lists are already sorted and each has $n = 15$ visitors, so in each case the median is the 8th value and the count is odd, meaning the median is left out of both halves.

(a) Layout 1. The median is the 8th value, 10 minutes. The lower half is 4, 6, 7, 7, 8, 9, 9, whose middle value is $Q_1 = 7$ minutes, and the upper half is 11, 12, 13, 14, 16, 18, 32, whose middle value is $Q_3 = 14$ minutes. The five-number summary is 4, 7, 10, 14, 32 minutes, and $IQR = 14 - 7 = 7$ minutes.

(a) Layout 2. The median is the 8th value, 16 minutes. The lower half is 9, 11, 12, 13, 14, 15, 15, whose middle value is $Q_1 = 13$ minutes, and the upper half is 17, 18, 19, 20, 21, 23, 25, whose middle value is $Q_3 = 20$ minutes. The five-number summary is 9, 13, 16, 20, 25 minutes, and $IQR = 20 - 13 = 7$ minutes.

(b) Both layouts have $IQR = 7$ minutes, so $1.5 \times IQR = 10.5$ minutes for each. Layout 1's fences are $7 - 10.5 = -3.5$ and $14 + 10.5 = 24.5$ minutes, and the 32-minute visit is above 24.5, so it is a high outlier. Layout 2's fences are $13 - 10.5 = 2.5$ and $20 + 10.5 = 30.5$ minutes, and its shortest visit of 9 minutes and longest of 25 minutes both sit inside, so Layout 2 has no outliers.

Means, needed for the shape check and for part (d). Layout 1 sums to 176 minutes, so $\bar{x} = 176/15 = 11.73$ minutes. Layout 2 sums to 248 minutes, so $\bar{x} = 248/15 = 16.53$ minutes.

(c) Shape. Layout 1's mean of 11.73 minutes sits 1.73 minutes above its median of 10 minutes, and its longest visit runs 22 minutes above the median while its shortest falls only 6 minutes below, so Layout 1 is skewed to the right. Layout 2's mean of 16.53 minutes sits within 0.53 minutes of its median of 16 minutes, and its two tails are close in length (7 minutes down, 9 minutes up), so Layout 2 is roughly symmetric.

(c) Comparison in context. Visitors stayed longer under Layout 2: its median of 16 minutes is 6 minutes above Layout 1's median of 10 minutes. The middle halves are spread identically, with $IQR = 7$ minutes for both, but Layout 1's full range of $32 - 4 = 28$ minutes is nearly twice Layout 2's $25 - 9 = 16$ minutes, and all of that extra width comes from a single visitor. Layout 1 is skewed to the right with a high outlier at 32 minutes, above its 24.5-minute fence, while Layout 2 is roughly symmetric with no outliers.

(d) Test the claim against both centers. Medians: $16/10 = 1.60$, so at the median a Layout 2 visit is 60% longer, matching the curator exactly. Means: $16.53/11.73 = 1.41$, so on the mean the increase is about 41%, not 60%. The claim holds only if the curator is talking about medians, and Layout 1's outlier is what drives the two figures apart by inflating that layout's mean.

(e) It blocks any claim that the layout caused the difference. Morning and afternoon visitors may differ in their characteristics or circumstances, so time of day changes alongside the layout and could account for the whole 6-minute gap on its own. These are also descriptive summaries of two groups of 15 visitors, not an inference, so nothing here supports a claim about the museum's visitors in general.