Statistics: questions with answers

The smallest reading is still $30$ and the largest is still $54$, so the range is still $24$. The range is built from those two readings alone; another reading can move without changing it provided it stays between those extremes. This is the limitation of the range: it says nothing about where the remaining readings sit or their distance from the mean.

Multiplying by $3$ multiplies every deviation by $3$ and every squared deviation by $3^2$, so the variance becomes $441$. Adding $3$ afterwards moves every observation and the mean together, so no deviation changes. The mean ends at $3 \times 10 + 3 = 33$ and the variance at $441$. Using $3$ times the old variance gives $147$, and adding the shift to the correct variance gives $444$. Wrongly scaling the standard deviation by $3^2$ gives $63$, and squaring that gives $3969$.

A mean is a measure of central tendency: it reports where the daily outputs are centred, and nothing else. From $60$ items a day over 15 days all that follows is that the outputs are centred at $60$ and that their total is $900$. Whether the daily outputs clustered near $60$ or swung far either side of it is exactly the question a measure of dispersion answers, and this report does not.

The five observations total $210$, so the mean is $\bar{x} = 42$. The deviations from the mean are 2, -14, 8, -6 and 10. Dropping the minus sign from each one separately gives the distances 2, 14, 8, 6 and 10, which total $40$. There are five observations, so $\text{M.D.}(\bar{x}) = \frac{40}{5} = 8$. Every observation counts, including those below the mean, and the divisor is the number of observations.

Sorted, the observations are 15, 16, 17, 19, 26 and 39, so the median is $\frac{17 + 19}{2} = 18$. The distances of the six observations from $18$ total $36$, so $\text{M.D.}(M) = \frac{36}{6} = 6$. The mean of these observations is a different number, $22$, and the mean deviation about it is $7$ - larger, as the chapter says it must be.

Mean deviation may be taken about any fixed value $a$: $\text{M.D.}(a) = \frac{\sum |x_i - a|}{n}$. Here the deviations from $65$ are -3, -6, -11, 18 and 27; dropping the minus signs one at a time gives distances totalling $65$, so $\text{M.D.}(65) = \frac{65}{5} = 13$. Note that the mean of these observations is $70$, not $65$, and the mean deviation about the mean is a different number, $14$.

Two of the chapter's stated drawbacks bear on a series like these waiting times. Where the degree of variability is very high the median is not a representative central tendency, so the mean deviation about it cannot be fully relied upon; and mean deviation, being built from absolute values, cannot be subjected to further algebraic treatment. The modulus can certainly be evaluated case by case, but it cannot be opened out and combined the way a square can, which is why a squares-based measure is introduced. Using only two observations is a limitation of the range, not of mean deviation.

First $\bar{y} = \frac{\sum f_i y_i}{N} = \frac{-80}{40} = -2$. The step deviations are a change of variable, so the mean is carried back with both operations undone: $\bar{x} = A + h\bar{y} = 65 + 5 \times -2 = 55$. Leaving the common factor out would give $63$, and offering $-2$ itself leaves the shift undone as well. The variance is different: shifting the origin changes no distance between observations, so $65$ never enters it and only the scale is restored, $\sigma_x^2 = h^2\sigma_y^2 = 50$. Adding the assumed mean there too would give $115$.

Line 2 has produced the squared deviations, so all that remains is Step 4 of the definition: add them and divide by the number of observations. Their total is $2000$ and there are five observations, so $\sigma^2 = \frac{2000}{5} = 400$. Stopping at the total $2000$ leaves a number that grows with the length of the list; averaging the distances instead of their squares gives $16$, which is the mean deviation; and averaging the squares of the readings themselves gives $4000$, which is not a measure of spread at all.

The variance is the square of the standard deviation, so the standard deviation is the positive square root of the variance: $\sigma = \sqrt{81} = 9$. Squaring the variance instead would give $6561$, which reverses the relation the symbol $\sigma^2$ already states. The unit also settles itself: the variance is in square centimetres because it is built from squared deviations, and taking the root brings the measure back to centimetres, the unit of the rods themselves.

$\sigma^2 = \frac{1}{n}\sum x_i^2 - (\bar{x})^2$, so the mean of the squares exceeds the square of the mean by exactly the variance: $130 - 81 = 49$. The standard deviation is the positive square root of that, $7$. The two quantities are equal only when the variance is zero, that is when every observation equals the mean, so a gap of $49$ is not an arithmetic slip. And $\sqrt{n\sum x_i^2 - (\sum x_i)^2} = \sqrt{1225} = 35$ is $n$ times the standard deviation, not the standard deviation itself.

Each class is represented by its mid-point: 35, 45, 55, 65 and 75. The frequencies 2, 8, 2, 4 and 4 balance about the middle mid-point, so $\bar{x} = 55$. Using the lower limits instead pulls every value down by half a class width and gives $50$. $45$ is the mid-point of $40-50$, the class with the most students, not the mean.

A frequency table is a shorthand for a longer list, so the number of observations is the total frequency $N = 1 + 1 + 3 + 9 + 6 = 20$, not the five values written in the top row. The cumulative frequencies are 1, 2, 5, 14 and 20, so the $10$th and $11$th observations - the middle two of the twenty - both fall in the fourth row, and the median is $28$. The middle row of the table, $20$, carries only three of the twenty observations, and the value halfway between the smallest and the largest, $24$, ignores the frequencies altogether.

The variance is the mean of the squared deviations over all the observations the table records. There are $N = 2 + 2 + 12 + 2 + 2 = 20$ of them, not five, and each squared deviation must be counted as many times as its own frequency. So the numerator is $18000$ and the denominator is $20$, giving $\sigma^2 = 900$. Using the five rows as the divisor applies half the weighting and drops the other half; using the unweighted total $9000$ drops the weighting altogether.

Omitting observations removes them from the data, so the total falls to $1200 - 96 = 1104$ and the count falls to $48$. The new mean is $\frac{1104}{48} = 23$. Both must fall together: dividing the reduced total by the original 50 treats the omitted observations as still present, and dividing the original total $1200$ by $48$ leaves the total unrepaired while the count has moved.