Single Variable Data

A linear change affects the mean exactly as it affects each value: new mean $= 2(12) + 3 = 27$. ($24$ forgot the $+3$; $15$ added 3 without doubling.)

With $n = 9$ the median is the 5th value. Changing only the three largest (positions 7, 8, 9) leaves the 5th value untouched, so the median stays $14$. (Unlike the mean, the median is resistant to changes in the extreme values.)

With $n = 8$, the lower half $\{4, 7, 8, 10\}$ gives $Q_1 = \dfrac{7 + 8}{2} = 7.5$ and the upper half $\{12, 15, 18, 21\}$ gives $Q_3 = \dfrac{15 + 18}{2} = 16.5$. So $\text{IQR} = 16.5 - 7.5 = 9$. ($17$ subtracted the wrong pair; $16.5$ is only $Q_3$.)

Multiplying every value by $3$ multiplies the spread by $3$, so the standard deviation becomes $3 \times 5 = 15$. Subtracting $4$ only shifts the data and leaves the spread unchanged. ($11$ wrongly subtracted 4 as well; $5$ ignored the multiplication.)

Standard deviation measures the typical distance from the mean ($30$). Every value in Set P is $20$ away from the mean, whereas Set Q's values cluster close to $30$. So Set P has the larger standard deviation, even though both share the same range. (Range uses only the two extremes; standard deviation uses every value.)

The mean condition gives $\dfrac{4 + 7 + x + 10 + 2x}{5} = 9$, so $21 + 3x = 45$, giving $3x = 24$ and $x = 8$. (Set up an equation from the mean, then solve it.)

When the median is pushed toward $Q_1$ and the upper whisker stretches out, the higher values are more spread out, forming a tail to the right, i.e. positively (right) skewed. (A symmetric set would have the median centred in the box with equal whiskers.)

Compare centre and spread together: Class A's median $72 > 70$ (typically higher), and its IQR $8 < 18$ (less spread, so more consistent). (A larger IQR means more spread, so Class B is the less consistent one.)

Here $Q_1 = 14$, $Q_3 = 18$, so $\text{IQR} = 4$. The upper fence is $Q_3 + 1.5 \times \text{IQR} = 18 + 6 = 24$, and $40 > 24$, so $40$ is an outlier. (The lower fence $14 - 6 = 8$ excludes nothing below.)

A percentile gives the percentage of the data lying below a value, so the $80$th percentile means about $80\%$ of students scored below Maya's mark (and about $20\%$ above). (It is not her exam score, nor her rank position.)

A box plot shows the five-number summary (minimum, $Q_1$, median, $Q_3$, maximum), so the median, range, IQR and maximum can all be read from it. It does not show how many values there are, so the number of students cannot be determined. (B to E are all readable from the plot.)

Class A IQR $= 80 - 60 = 20$; Class B IQR $= 78 - 65 = 13$. So Class A has the larger IQR. (Spread comes from $Q_3 - Q_1$, not the mean.)

First find the mean: $\dfrac{3 + 5 + 5 + 8 + 9 + 12 + 15}{7} = \dfrac{57}{7} \approx 8.1$. The values above it are $9,\ 12,\ 15$ (three of the seven), so the probability is $\dfrac{3}{7}$. (Find the mean, then apply probability.)

Weight each mean by its share of the total: using $2$ and $3$ parts, $\dfrac{2(60) + 3(85)}{2 + 3} = \dfrac{120 + 255}{5} = \dfrac{375}{5} = 75$. ($72.5$ is the plain average of $60$ and $85$, which ignores the ratio of group sizes.)

Median $= \dfrac{7 + 9}{2} = 8$; mean $= \dfrac{135}{6} = 22.5$. The outlier $100$ pulls the mean far above the median, so the mean is much greater. (The median barely moves.)

The mean is $\dfrac{40}{8} = 5$. The squared deviations are $9, 1, 1, 1, 0, 0, 4, 16$, summing to $32$, so the variance is $\dfrac{32}{8} = 4$ and the standard deviation is $\sqrt{4} = 2$. ($4$ is the variance; $32$ is the sum of squared deviations.)

Increasing every value by $20\%$ multiplies each one by $1.2$. Multiplying by a constant scales both the mean and the spread by that constant, so both are $\times 1.2$. (Adding a constant would shift the mean but leave the SD unchanged; a percentage increase is a multiplication, not an addition.)

Most households earn modest amounts while a few earn very large amounts, creating a long right tail, so income is positively skewed. (Heights and shoe sizes are roughly symmetric; a fair die is uniform; an easy test is negatively skewed, with a tail toward the low scores.)

For an odd count of consecutive whole numbers the mean is the middle value, so the middle number is $18$: the five numbers are $16, 17, 18, 19, 20$ and the largest is $20$. (Symmetry of consecutive numbers avoids having to solve $\tfrac{5n + 10}{5} = 18$.)

With $40$ students the median is around the $20$th–$21$st value. The cumulative frequency reaches $17$ by the end of the 11–20 interval and $32$ by the end of the 21–30 interval, so the middle values fall in the 21–30 interval.

$\text{mean} = \dfrac{2 \times 3 + 5 \times 4 + 8 \times 3}{3 + 4 + 3} = \dfrac{6 + 20 + 24}{10} = \dfrac{50}{10} = 5$. ($50$ is the total without dividing; $15$ added the values ignoring frequency.)

For an even count the median is the average of the two middle values: $\dfrac{9 + x}{2} = 11$, so $9 + x = 22$ and $x = 13$. ($22$ is the total of the two values, not $x$.)

Old total $= 6 \times 15 = 90$; new total $= 7 \times 16 = 112$. The 7th number is $112 - 90 = 22$. ($16$ is the new mean, not the added value.)

The median is the middle (3rd) value, $x$, and the mean is $\dfrac{4 + 8 + x + 14 + 14}{5} = \dfrac{40 + x}{5}$. Setting them equal: $x = \dfrac{40 + x}{5}$, so $5x = 40 + x$, giving $4x = 40$ and $x = 10$. (Check: $4, 8, 10, 14, 14$ has both mean and median $10$.)

The variance is the mean of the squared deviations: $\dfrac{80}{5} = 16$. The standard deviation is $\sqrt{16} = 4$. ($16$ is the variance, not the SD; $\sqrt{80}$ forgot to divide by $5$ first.)