Statistics (Mean, Median, Mode) Questions and Answers

Answer: B

She got 3 the maximum number of times.

Counting each result in 5, 3, 6, 6, 1, 4, 5, 3, 3, 2:

• 3 appears three times.
• 5 and 6 appear twice each.
• 1, 2 and 4 appear once each.

So 3 is the most frequent result, the mode of the data. The options 6 and 5 occur only twice, and 1 only once.

Answer: C

The difference between y and x is 3.

Arrange the original data in order: 3, 7, 8, 15, 17, 21, 23, 24, 29, 30. With ten values the median is the mean of the 5th and 6th: x = (17 + 21)/2 = 19.

After replacing 3 by 33 and 8 by 18 the data are 7, 15, 17, 18, 21, 23, 24, 29, 30, 33, so y = (21 + 23)/2 = 22.

\[ y - x = 22 - 19 = 3 \]

Forgetting to re-sort the data after the replacement is the usual source of the other answers.

Answer: C

The sum of x and y is 29.

There are 15 values, so the median is the 8th value in order.

Original data in order: 8, 9, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16, 16, 17, 17. The 8th value is 14, so x = 14.

Replacing 8 by 18: 9, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16, 16, 17, 17, 18. The 8th value is now 15, so y = 15.

x + y = 14 + 15 = 29. 28 would follow if the median did not change, which forgets that removing the smallest value shifts every position by one.

Answer: C

The difference is 5.5.

\[ \begin{aligned} \text{Sum} &= 366, \quad n = 12 \\ \text{Mean} &= 366 / 12 = 30.5 \\ \text{Range} &= 42 - 17 = 25 \\ 30.5 - 25 &= 5.5 \end{aligned} \]

The range is the highest value minus the lowest, 42 and 17. Using 18 as the lowest, or miscounting the 12 values, gives the other options.

Answer: A

The median is 13.5.

Arrange the ten numbers in order:

9, 10, 11, 11, 13, 14, 15, 17, 18, 69

With an even count, the median is the mean of the two middle values, the 5th and 6th: \((13 + 14)/2 = 13.5\).

14 takes only the 6th value, and 11 is the mode (it appears twice), not the median. 18.7 is the mean, \(187/10\), pulled up by the outlier 69; the median is preferred here exactly because it ignores such an extreme value.

Answer: A

The median is 145.0.

Arrange the ten numbers in order: 130, 137, 138, 138, 144, 146, 149, 154, 156, 160. With an even count, the median is the mean of the 5th and 6th values:

\[ \text{median} = \frac{144 + 146}{2} = 145.0 \]

145.2 is the arithmetic mean of the set (1452 divided by 10), not the median. 151.0 is the mean of 146 and 156, values taken from the unsorted list, and 146.0 is just the 6th value on its own.

Answer: C

The statement that must be true is At most half the candidates have a score that is larger than the median.

By definition, the median splits the ordered scores so that at least half are at or below it and at least half are at or above it. So no more than half can lie strictly above it, whatever the shape of the distribution.

The mean and median can be in either order: a distribution skewed to the right has its mean above the median, one skewed to the left has its mean below it. So neither 'median at least the mean' nor 'mean at least the median' must hold. And far more than half the candidates can score above the mean, if a few very low scores pull the mean down.

Answer: B

The median salary is Rs. 30,000.

With 100 employees the median is the average of the 50th and 51st salaries in order. The first 45 earn Rs. 20,000, and employees 46 to 70 earn Rs. 30,000, so both the 50th and 51st earn Rs. 30,000.

Rs. 20,000 is the mode (the most common salary). Rs. 32,300 is close to the mean, which is pulled up by the two very high salaries: (45 × 20 + 25 × 30 + 20 × 40 + 8 × 60 + 2 × 150) thousand / 100 = Rs. 32,300. Rs. 40,000 is further up the list.

Answer: C

The correct assertions are P and R.

• P is true: adding a constant to every entry adds it to the mean.
• Q is false: adding a constant shifts every entry and the mean equally, so the deviations from the mean, and the standard deviation, are unchanged.
• R is true: doubling every entry doubles the mean.
• S is false: doubling every entry doubles every deviation, so the standard deviation doubles.

Every other option includes Q or S, so P, R is the only fully correct choice.

Answer: B

The correct order is mode \(\gt\) median \(\gt\) mean.

There are \(3 + 9 + 11 + 7 + 14 + 2 + 4 = 50\) students.

• Mode: the highest frequency, 14, is at 7 marks.
• Median: the 25th and 26th values. Cumulative frequencies are 3, 12, 23, 30, ..., so both fall at 6 marks: median 6.
• Mean: \((9 + 36 + 55 + 42 + 98 + 16 + 36)/50 = 292/50 = 5.84\).

So 7 \(\gt\) 6 \(\gt\) 5.84. The other options put the mean above the median or the median above the mode, which these numbers contradict.