Choosing mean vs median and SD vs IQR

Sort the heights: 19, 20, 21, 22, 23, 23, 23, 24, 25, 26, 27. There are $n = 11$ seedlings.

(a) Mean: the heights sum to 253, so $\bar{x} = 253/11 = 23$ cm. Median: with 11 values it is the 6th, which is 23 cm.

(a) Quartiles, median excluded. The lower half is 19, 20, 21, 22, 23, so $Q_1 = 21$ cm. The upper half is 23, 24, 25, 26, 27, so $Q_3 = 25$ cm. Then $IQR = 25 - 21 = 4$ cm.

(a) Standard deviation: the deviations from 23 are $-4, -3, -2, -1, 0, 0, 0, 1, 2, 3, 4$, whose squares sum to 60. So $s^2 = 60/10 = 6$ and $s = \sqrt{6} \approx 2.45$ cm.

(b) Run the outlier check first: $1.5 \times 4 = 6$ cm, so the fences are $21 - 6 = 15$ cm and $25 + 6 = 31$ cm. The minimum of 19 and the maximum of 27 both sit well inside, so nothing is flagged.

(b) Now look at the sorted list. It falls away from 23 by roughly equal amounts on both sides, and the mean and the median agree exactly at 23 cm, so nothing is dragging the mean off center.

(b) That is the case where the mean and the standard deviation earn their keep, because both use every value. As a check that $s$ describes a typical distance, 7 of the 11 heights lie within $\sqrt{6} \approx 2.45$ cm of the mean.

Sort the gifts: 10, 12, 15, 15, 18, 20, 25, 30, 40, 275. There are $n = 10$ gifts.

(a) Mean: the gifts sum to 460, so $\bar{x} = 460/10 = 46$ dollars. Median: with an even count it is the average of the 5th and 6th values, $(18 + 20)/2 = 19$ dollars.

(a) Quartiles. The count is even, so the lower half is the first five gifts 10, 12, 15, 15, 18, giving $Q_1 = 15$ dollars, and the upper half is 20, 25, 30, 40, 275, giving $Q_3 = 30$ dollars. Then $IQR = 30 - 15 = 15$ dollars.

(a) Standard deviation: the squared deviations from 46 sum to 59,008, so $s^2 = 59{,}008/9 \approx 6556.44$ and $s \approx 80.97$ dollars.

(b) Outlier check: $1.5 \times 15 = 22.5$ dollars, so the fences are $15 - 22.5 = -7.5$ dollars and $30 + 22.5 = 52.5$ dollars. The 275-dollar gift is far beyond the upper fence, so it is an outlier.

(b) Look at what the outlier did to each summary. The mean of 46 dollars is larger than 9 of the 10 gifts, so it describes none of the donors. The standard deviation of about 80.97 dollars is larger than the entire range of the other nine gifts, which run from 10 to 40 dollars.

(b) The median of 19 dollars and the IQR of 15 dollars barely notice the 275-dollar gift, because the median depends on position and the IQR on where the middle half sits. Report those two, and mention the large gift separately rather than letting it hide inside an average.

Sort the commutes: 6, 7, 8, 8, 9, 41, 42, 43, 44, 45. There are $n = 10$ employees.

(a) Mean: the commutes sum to 253, so $\bar{x} = 253/10 = 25.3$ minutes. Median: the average of the 5th and 6th values, $(9 + 41)/2 = 25$ minutes. The two are 0.3 minutes apart, exactly as the manager says.

(b) Now read the sorted list. Five commutes are 9 minutes or less, five are 41 minutes or more, and nothing lies between 9 and 41 minutes.

(b) Measure the distance from the center to the nearest real value. The closest commute to the mean is the 41-minute one, and it sits $41 - 25.3 = 15.7$ minutes away. No employee commutes anything close to 25 minutes, so the center describes nobody.

(b) The outlier rule does not rescue the argument either. $Q_1$ is the median of 6, 7, 8, 8, 9, which is 8 minutes, and $Q_3$ is the median of 41, 42, 43, 44, 45, which is 43 minutes, so $IQR = 35$ minutes and the fences sit at $8 - 52.5 = -44.5$ and $43 + 52.5 = 95.5$ minutes. Nothing is flagged. Passing the outlier check is not evidence that a mean is safe.

(b) For the record, $s \approx 18.70$ minutes here, which is smaller than the IQR of 35 minutes. Neither number is a typical distance from a center, because there is no typical commute.

(c) The honest summary starts with shape. This distribution has two separated clusters, one from 6 to 9 minutes and one from 41 to 45 minutes, which is what a report should say, with a center and a spread given for each cluster rather than one center for all 10 employees.

(a) Sort the 8 loaves: 31, 33, 34, 35, 36, 37, 38, 40. They sum to 284, so $\bar{x} = 284/8 = 35.5$ minutes, and the median is $(35 + 36)/2 = 35.5$ minutes.

(a) Quartiles: the lower half is 31, 33, 34, 35, so $Q_1 = (33 + 34)/2 = 33.5$ minutes, and the upper half is 36, 37, 38, 40, so $Q_3 = (37 + 38)/2 = 37.5$ minutes. Then $IQR = 37.5 - 33.5 = 4$ minutes.

(a) Standard deviation: the squared deviations from 35.5 sum to 58, so $s^2 = 58/7 \approx 8.29$ and $s = \sqrt{58/7} \approx 2.88$ minutes.

(b) The prediction: the mean and the standard deviation should move most, because both use the value 94 itself, while the median and the IQR only care about where values sit in order.

(b) With the ninth loaf the sorted list is 31, 33, 34, 35, 36, 37, 38, 40, 94, summing to 378. New mean: $378/9 = 42$ minutes. New median: with 9 values it is the 5th, which is 36 minutes.

(b) New quartiles. The count is odd, so leave the median out. The lower half is 31, 33, 34, 35, giving $Q_1 = 33.5$ minutes, and the upper half is 37, 38, 40, 94, giving $Q_3 = (38 + 40)/2 = 39$ minutes. Then $IQR = 39 - 33.5 = 5.5$ minutes.

(b) New standard deviation: the squared deviations from 42 sum to 3100, so $s^2 = 3100/8 = 387.5$ and $s = \sqrt{387.5} \approx 19.69$ minutes.

(c) Line up the four moves. The mean rose 6.5 minutes while the median rose 0.5 minutes, so the mean moved 13 times as far. The standard deviation went from about 2.88 to about 19.69 minutes, roughly 6.8 times larger, while the IQR went from 4 to 5.5 minutes.

(c) Note that the IQR did move. Resistant means barely moved, not frozen: adding a ninth value changes which values the quartiles land on. Note also that the new mean of 42 minutes is longer than 8 of the 9 cooling times, and that the 94-minute loaf is an outlier, since the new fences are $33.5 - 8.25 = 25.25$ and $39 + 8.25 = 47.25$ minutes.

Sort the original masses: 24, 27, 29, 31, 32, 34, 36, 37, 38. There are $n = 9$ cores.

Prediction: the corrected value stays the smallest in the list, so no other value changes position. The median and the quartiles read positions, so they should hold; the mean and the standard deviation read values, so they should move.

Before, the masses sum to 288, so $\bar{x} = 288/9 = 32$ grams, and the median is the 5th value, 32 grams.

Before, the lower half is 24, 27, 29, 31, so $Q_1 = (27 + 29)/2 = 28$ grams; the upper half is 34, 36, 37, 38, so $Q_3 = (36 + 37)/2 = 36.5$ grams; and $IQR = 36.5 - 28 = 8.5$ grams.

Before, the squared deviations from 32 sum to 180, so $s^2 = 180/8 = 22.5$ and $s = \sqrt{22.5} \approx 4.74$ grams.

After the correction the sorted list is 6, 27, 29, 31, 32, 34, 36, 37, 38. The 5th value is still 32 grams, the lower half is still four values (6, 27, 29, 31) whose median is still 28 grams, and the upper half has not changed at all. So the median stays 32, $Q_1$ stays 28, $Q_3$ stays 36.5, and the IQR stays 8.5 grams.

The mean does move, and by a predictable amount: the total fell by $24 - 6 = 18$ grams spread over 9 cores, so the mean falls $18/9 = 2$ grams, from 32 to 30 grams. Check it directly: the new total is 270 and $270/9 = 30$ grams.

The standard deviation moves more. The squared deviations from 30 sum to 756, so $s^2 = 756/8 = 94.5$ and $s = \sqrt{94.5} \approx 9.72$ grams, about 2.05 times its former value.

One more consequence. The fences are $28 - 12.75 = 15.25$ grams and $36.5 + 12.75 = 49.25$ grams in both versions, because the quartiles never moved. The 24-gram reading sat inside them; the corrected 6-gram core sits below 15.25, so it is now an outlier.

Sort the salaries: 32,000, 34,000, 35,000, 36,000, 38,000, 39,000, 41,000, 44,000, 196,000. There are $n = 9$ employees.

Compute both centers. The salaries sum to 495,000, so $\bar{x} = 495{,}000/9 = 55{,}000$ dollars, and the median is the 5th value, 38,000 dollars.

Check the shape with the quartiles. The lower half is 32,000, 34,000, 35,000, 36,000, so $Q_1 = 34{,}500$ dollars; the upper half is 39,000, 41,000, 44,000, 196,000, so $Q_3 = (41{,}000 + 44{,}000)/2 = 42{,}500$ dollars; and $IQR = 8{,}000$ dollars. The upper fence is $42{,}500 + 12{,}000 = 54{,}500$ dollars, so the owner's salary is an outlier.

(a) The mean of 55,000 dollars is above 8 of the 9 salaries, and it even sits above the 54,500-dollar outlier fence, so it describes no one at the studio. The median of 38,000 dollars is an actual salary with four people above it and four below. Report the median for a typical earner, with the IQR of 8,000 dollars for spread.

(b) The budget needs a total, and the total is exactly $n$ times the mean: $9 \times 55{,}000 = 495{,}000$ dollars, which matches the sum of the nine salaries.

(b) The median cannot do that job. Nine salaries at the median would be $9 \times 38{,}000 = 342{,}000$ dollars, which understates the real payroll by $495{,}000 - 342{,}000 = 153{,}000$ dollars.

The shape of the data did not change between the two parts, only the question did. Skew rules the mean out as a description of a typical value, and it does not rule the mean out when the quantity you need is a total.

(a) Compare the mean with the quartiles. The mean of 6.4 days sits above $Q_3 = 6$ days, so about three quarters of the admissions were shorter than the average stay, and the mean is more than double the median of 3 days.

(a) A summary that most of the data falls below is a poor answer to "typical". The reporter should use the median of 3 days, with the IQR for spread.

(b) $IQR = Q_3 - Q_1 = 6 - 2 = 4$ days, so $1.5 \times IQR = 6$ days.

(b) Upper fence: $6 + 6 = 12$ days. The maximum of 96 days is far beyond it, so at least one stay is an outlier. Lower fence: $2 - 6 = -4$ days. A length of stay cannot be negative and the minimum is 1 day, so no stay can fall below the lower fence and nothing can be flagged at the low end.

(c) Settled: the long tail runs to the right. The maximum sits 90 days above $Q_3$ while the minimum sits only 1 day below $Q_1$, the distance from the median up to $Q_3$ (3 days) is three times the distance from $Q_1$ up to the median (1 day), and the mean has been pulled past $Q_3$.

(c) The standard deviation fits that story. At 7.9 days it is nearly double the IQR of 4 days, because $s$ counts the 96-day stay in full while the IQR only reports where the middle half sits, and part (b) already showed that the extreme values can only be at the high end.

(c) Left open: how many peaks the distribution has. Summary statistics cannot show clustering, which is the trap in problem 3, where a mean and a median 0.3 minutes apart sat in an empty gap between two groups. A histogram or a boxplot is needed before claiming the distribution has a single peak.

Sort the counts: 22, 24, 25, 25, 27, 28, 29, 30, 31, 33, 34, 118. There are $n = 12$ posts.

(a) Mean: the counts sum to 426, so $\bar{x} = 426/12 = 35.5$ comments. Median: the average of the 6th and 7th values, $(28 + 29)/2 = 28.5$ comments.

(a) Quartiles. The count is even, so the lower half is 22, 24, 25, 25, 27, 28, giving $Q_1 = (25 + 25)/2 = 25$ comments, and the upper half is 29, 30, 31, 33, 34, 118, giving $Q_3 = (31 + 33)/2 = 32$ comments. Then $IQR = 32 - 25 = 7$ comments.

(a) Standard deviation: the squared deviations from 35.5 sum to 7571, so $s^2 = 7571/11 \approx 688.27$ and $s \approx 26.24$ comments.

(b) Outlier check: $1.5 \times 7 = 10.5$, so the fences are $25 - 10.5 = 14.5$ and $32 + 10.5 = 42.5$ comments, and the 118-comment post is an outlier. The mean of 35.5 is above 11 of the 12 counts and above $Q_3$, and $s \approx 26.24$ is larger than the gap between the smallest and largest of the other eleven posts, which run from 22 to 34.

(b) So report the median of 28.5 comments and the IQR of 7 comments, and mention the newsletter post on its own.

(c) Drop the 118-comment post. The remaining 11 counts sum to $426 - 118 = 308$, so $\bar{x} = 308/11 = 28$ comments, and the median is now the 6th value, 28 comments.

(c) New quartiles: the lower half is 22, 24, 25, 25, 27, giving $Q_1 = 25$, and the upper half is 29, 30, 31, 33, 34, giving $Q_3 = 31$, so $IQR = 6$ comments. New standard deviation: there are 11 posts now, so the divisor is 10. The squared deviations from 28 sum to 146, so $s^2 = 146/10 = 14.6$ and $s = \sqrt{14.6} \approx 3.82$ comments.

(c) The moves: the mean fell 7.5 comments (35.5 to 28) against the median's 0.5 (28.5 to 28), and $s$ fell from about 26.24 to about 3.82 comments while the IQR fell by 1 comment (7 to 6). The mean and the median now agree exactly, and no count is flagged by the new fences of 16 and 40 comments.

(d) The claim fails. In problem 3 the mean of 25.3 minutes and the median of 25 minutes are 0.3 minutes apart, yet the commutes split into a cluster near 6 to 9 minutes and a cluster near 41 to 45 minutes, with the nearest commute 15.7 minutes from the mean. Close centers are consistent with symmetry, not proof of it, so graph the data before trusting the mean and $s$.