Data

Mean $= \dfrac{3 + 5 + 7 + 11 + 4}{5} = \dfrac{30}{5} = 6$. (5 is the median; 8 is the range; 30 is the sum.)

Sort first: 1, 3, 5, 8, 10. The middle (3rd) value is 5. (5.4 is the mean; not sorting gives the wrong 8.)

Sort: 2, 4, 6, 7, 9, 11. With 6 values, the median is the average of the two middle values: $\dfrac{6 + 7}{2} = 6.5$.

5 appears three times, more than any other value, so the mode is 5. (3 is its frequency, not the mode.)

Range $= 22 - 4 = 18$. (Adding gives 26; range measures spread, not a typical value.)

Each leaf combines with the stem: stem 2, leaf 3 is 23, and so on, giving 23, 25, 25, 27, 28. (Listing 2, 3, 5, ... treats the stem and leaves as separate values.)

The mode is the value with the most students. 1 pet has 7 students, more than any other, so the mode is 1 pet. (7 is the frequency, not the value.)

Scores under 20 are the first two bars: $2 + 5 = 7$ students. Total $= 2 + 5 + 8 + 4 + 1 = 20$, so the fraction is $\dfrac{7}{20}$. ($\dfrac{13}{20}$ is the complement, who scored 20 or more.)

The modal class has the highest frequency: 30–39 with 12 students. (12 is the frequency; 50–59 is just the highest interval, not the most common.)

When the tail extends to the right (most data on the left), the distribution is positively (right) skewed. The skew is named after the side the tail is on.

The mean adds in every value, so one very large value pulls it upward, while the median (the middle position) is barely affected. So the mean is usually larger than the median for a right-skewed dataset.

Starting the axis above 0 (a truncated axis) makes small differences between columns look much larger than they really are. It is a common way graphs mislead, even though the data itself is unchanged.

A frequency histogram groups the data into intervals (classes) and shows how many values fall in each, making the overall spread and shape easy to see. A single number like the mean hides the spread.

Total goals $= 0\times 2 + 1\times 5 + 2\times 3 = 0 + 5 + 6 = 11$; total games $= 2 + 5 + 3 = 10$; mean $= \dfrac{11}{10} = 1.1$.

With an even number of values there is no single middle value, so the median is the average of the two middle ones, which are the 4th and 5th values. (Averaging all 8 gives the mean, not the median.)

Range of A $= 9 - 5 = 4$; range of B $= 13 - 1 = 12$. Both have mean 7, but B is more spread out. The mean alone does not show spread.

Old mean $= \dfrac{420}{5} = 84$; new mean $= \dfrac{420 + 30}{6} = \dfrac{450}{6} = 75$. Adding a value below the mean makes it decrease.

The median is robust to outliers (it depends only on the middle position), while the mean is pulled towards an extreme value. So the median is the better measure of a typical value.

Weighted total $= 5\times 6 + 8\times 7 + 4\times 8 + 3\times 9 = 30 + 56 + 32 + 27 = 145$; total students $= 20$; mean $= \dfrac{145}{20} = 7.25$. (Averaging the distinct marks gives the wrong 7.5.)

The range measures spread. A larger range for Class X means its marks are more spread out (more variable), even though the typical mark (median) is the same for both classes.

The sum must be $5\times 8 = 40$. The four known numbers sum to $5 + 9 + 11 + 7 = 32$, so the fifth is $40 - 32 = 8$.

Total $= 7\times 3 + 8\times 5 + 9\times 2 = 21 + 40 + 18 = 79$; number of pairs $= 3 + 5 + 2 = 10$; mean $= \dfrac{79}{10} = 7.9$. (Averaging the three sizes gives the wrong 8.)

Original sum $= 6\times 10 = 60$. Removing 10 leaves a sum of $50$ over 5 values: $\dfrac{50}{5} = 10$. Removing a value equal to the mean does not change the mean.

The mean uses every value, so one extreme value shifts it noticeably. The median (middle position) and mode (most frequent value) are barely affected by a single outlier.

Old sum $= 4\times 6 = 24$; new sum $= 24 + 11 = 35$; new mean $= \dfrac{35}{5} = 7$. (Averaging 6 and 11 directly is wrong, since 6 already represents 4 values.)