Statistics: Box Plots and IQR

Sort first: $3, 5, 7, 8, 9, 11, 12$. With 7 values (odd), the median is the 4th (middle) value: 8. (7.86 is the mean; 12 is the maximum; 9 is one position off.)

Sort: $2, 4, 6, 7, 9, 11$. With 6 values (even), average the two middle values: $\dfrac{6 + 7}{2} = 6.5$. (6 forgot to average; 13 added without dividing; 6.83 is the mean.)

The lower half (the 4 values below the median 7.5) is $2, 4, 5, 7$, so $Q_1 = \dfrac{4 + 5}{2} = 4.5$. (7.5 is the median; 2 is the minimum.)

The upper half is $8, 10, 11, 13$, so $Q_3 = \dfrac{10 + 11}{2} = 10.5$. (13 is the maximum; 7.5 is the median.)

The IQR is $Q_3 - Q_1$, the spread of the middle 50% of the data. Because it uses only the quartiles, extreme values (outliers) do not affect it, which makes it a resistant measure of spread. (The difference between the largest and smallest values is the range.)

The quartiles enclose the middle 50% of the data (this is exactly what the box of a box plot shows), so about $50\%$ of $80 = 40$ values lie between $Q_1$ and $Q_3$. ($20$ is 25%, a single quarter; $80$ is all of the data.)

7 values, so median = 4th = 14. Lower half $5, 8, 12$ gives $Q_1 = 8$; upper half $18, 20, 24$ gives $Q_3 = 20$. Summary: 5, 8, 14, 20, 24. (Using 12 as the median or wrong quartile positions are the common slips.)

The five numbers total $5 \times 12 = 60$. The four known values sum to $8 + 10 + 14 + 16 = 48$, so the fifth is $60 - 48 = 12$. ($48$ is the sum of the four known values; $60$ is the total of all five.)

$IQR = Q_3 - Q_1 = 30 - 15 = 15$. (28 is the range $38 - 10$; 22 is the median; 45 added the quartiles.)

The box spans $Q_1$ to $Q_3$, the middle 50% of the data. (25% describes a single quartile; 100% would include the whiskers as well.)

The median is the line inside the box, at 22. (15 is $Q_1$ (left side of the box); 30 is $Q_3$ (right side).)

$Q_1$ is the left side of the box, at 15. (22 is the median; 10 is the minimum (end of the left whisker); 30 is $Q_3$.)

Work with totals, not the average of the averages: $20 \times 70 + 10 \times 85 = 1400 + 850 = 2250$ marks over $30$ students, so the mean is $\dfrac{2250}{30} = 75$. ($77.5$ is the plain average of $70$ and $85$, which ignores the different class sizes.)

The upper quartile $Q_3$ is the 75th percentile, so $75\%$ of the data lies below it and $25\%$ lies above. The probability is $0.25$. ($0.5$ is the probability of being above the median.)

$IQR = Q_3 - Q_1 = 30 - 15 = 15$, the width of the box. (28 is the range (full whisker-to-whisker spread); 22 is the median.)

Class A's box is wider: $IQR_A = 85 - 55 = 30$ versus $IQR_B = 80 - 65 = 15$. Both share the same median (70), but Class A is more spread out. (Class B has the smaller IQR; the ranges differ too, 55 versus 35.)

Skew is named for the long-tail direction. A long right whisker (with the median bunched toward $Q_1$) means the distribution is skewed right (positive). (A long left whisker would be skewed left; symmetric has equal whiskers and a centred median; bimodal cannot be read from a box plot.)

$IQR = 30 - 10 = 20$. Lower bound $= 10 - 1.5 \times 20 = -20$; upper bound $= 30 + 1.5 \times 20 = 60$. A value outside $[-20, 60]$ is an outlier, so $65 > 60$ is the outlier. (35 and 50 are above $Q_3$ but still within 60; being above $Q_3$ alone is not enough.)

The median depends only on the middle value(s), so it is resistant to outliers. The mean is sensitive because every value contributes to the sum. (Range and maximum involve the extreme value directly.)

$Q_1 = 60$ is the 25th percentile, so about 25% of students scored below 60. (50% is below the median; 75% is below $Q_3$; 60% and 40% confuse a value with a percentage.)

For symmetric data the quartiles are equidistant from the median, each $\dfrac{30}{2} = 15$ away: $Q_1 = 60 - 15 = 45$ and $Q_3 = 60 + 15 = 75$. (Q1 = 50, Q3 = 70 gives a difference of 20, not 30.)

The new sorted set $3, 7, 10, 15, 20, 100$ has 6 values, so the median is $\dfrac{10 + 15}{2} = 12.5$. (25.92 is the new mean; note the mean jumps from 11 to 25.92 but the median only moves to 12.5, since the median is resistant to outliers.)

With 4 values (even), the median is the average of the two middle values (the 2nd and 3rd): $\dfrac{8 + x}{2} = 9$. Solving: $8 + x = 18$, so $x = 10$. (9 used the median value itself as $x$; 18 stopped at $8 + x = 18$ without solving; 8.5 averaged 8 and 9.)

A larger IQR means more spread, so Class B (IQR 25) has more variable scores than Class A (IQR 15). The medians are equal (70), so neither scored higher in the centre. (Same median but different spread is the key idea.)

$IQR = 30 - 20 = 10$. Lower boundary $= Q_1 - 1.5 \times IQR = 20 - 15 = 5$. (15 is $1.5 \times IQR$ alone (the buffer); 45 is the upper boundary $Q_3 + 1.5 \times IQR$.)