Statistics (Mean, Median, Mode) Questions and Answers

Answer: B

Incorrect sum = Mean × Count = 27 × 5 = 135.

The difference due to misreading = Correct value - Incorrect value = 72 - 27 = 45.

The correct sum is 45 more than the calculated sum.

Correct sum = 135 + 45 = 180.

Correct mean = \(\frac{180}{5} = 36\).

Answer: A

The mean of the three measures is 13.

In order the data are 7, 8, 9, 10, 11, 11, 13, 14, 15, 16, 17, 23 (12 values).

\[ \begin{aligned} \text{Mode} &= 11 \\ \text{Median} &= \frac{11 + 13}{2} = 12 \\ \text{Range} &= 23 - 7 = 16 \\ \text{Mean} &= \frac{11 + 12 + 16}{3} = 13 \end{aligned} \]

12 is the median itself, not the mean of the three measures. 12.5 and 13.5 do not arise from the correct mode, median and range; the usual slips are forgetting to sort the data or to average the two middle values for the median.

Answer: C

The difference is \(\frac{8}{11}\).

The sum of the 11 values is 1 + 0 + 2 + 4 + 0 + 3 + 1 + 1 + 2 + 4 + 1 = 19, and 1 occurs most often (four times), so the mode is 1.

\[ \begin{aligned} \text{mean} &= \frac{19}{11} \\ \text{mean} - \text{mode} &= \frac{19}{11} - 1 = \frac{8}{11} \end{aligned} \]

Fractions with denominator 9 cannot arise from 11 values, and \(\frac{7}{11}\) would need the wickets to total 18.

Answer: B

The median is 54.

Use the mean to find x. Seven observations with mean 53 have a total of 7 × 53 = 371:

\[ \begin{aligned} 42 + 48 + 3x + 52 + 58 + 60 &= 371 \\ 260 + 3x &= 371,\ x = 37 \end{aligned} \]

The data are 42, 48, 52, 54, 57, 58, 60. With seven values in ascending order, the median is the fourth, x + 17 = 54. Option A, 53, is the mean, not the median.

Answer: A

The median is 75.

The median is the middle value when the data are arranged in order. The five marks are 75, 75, 75, 75, 75, so the middle (third) value is 75; when every value is the same, the mean, median and mode all equal that value.

375 is the total of the marks, 15 is 75 divided by 5, and 70 has no connection with the data.

Answer: B

The range is 16 °C.

The range of a data set is the highest value minus the lowest value. Here the highest temperature is 8 °C and the lowest is −8 °C:

\[ 8 - (-8) = 8 + 8 = 16 \]

Subtracting a negative number adds its size, which is the step the distractors get wrong. 0 °C comes from ignoring the minus sign (8 − 8), and 10 °C from taking −2 as the lowest value; 18 °C does not follow from the data.

Answer: C

The mean is 33.

Arrange the 17 values in order:

0, 8, 9, 11, 11, 17, 18, 20, 24, 25, 25, 25, 29, 30, 30, 44, 50

The median is the 9th value, 24. The mode is 25 (it occurs three times). The range is 50 − 0 = 50.

\[ \text{mean} = \frac{24 + 25 + 50}{3} = \frac{99}{3} = 33 \]

Taking the median from the unsorted list (the 9th value there is 30) or the range as 50 − 8 = 42 leads to the other options.

Answer: B

The difference is 2·5.

Arrange the 14 observations in order:

12, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 18, 19, 19

With an even number of values, the median is the mean of the 7th and 8th:

\[ \begin{aligned} \text{Median} &= \frac{14 + 15}{2} = 14.5 \\ \text{Mode} &= 12 \\ 14.5 - 12 &= 2.5 \end{aligned} \]

12 is the mode because it occurs three times, more than any other value. Reading the middle values from the unsorted list instead of the sorted one is the usual error, and it does not give 2·5.

Answer: B

The mean of the median and mode is 53.

Arrange the ten observations in order: 20, 26, 28, 30, 45, 55, 56, 56, 60, 61.

\[ \begin{aligned} \text{median} &= \frac{45 + 55}{2} = 50 \\ \text{mode} &= 56 \\ \text{mean of the two} &= \frac{50 + 56}{2} = 53 \end{aligned} \]

With an even number of values the median is the average of the 5th and 6th values. Taking 55 alone as the median gives 55.5, and the mode alone is 56, which is why 55 and 56 are traps.

Answer: B

The mean is 59.3.

There are 10 observations, so the median is the average of the 5th and 6th, which are a and a + 3.

\[ \begin{aligned} \frac{a + (a + 3)}{2} &= 61.5 \\ 2a + 3 &= 123 \\ a &= 60 \end{aligned} \]

So the data are 28, 31, 47, 51, 60, 63, 71, 74, 78, 90, whose sum is 593.

\[ \text{Mean} = \frac{593}{10} = 59.3 \]

Forgetting the + 3 in the sixth value gives a sum of 590 and a mean of 59, which is not offered; none of the other options fits a = 60.