Mean, Median and Mode: Practice problems with answers

Sorted data: 2, 3, 4, 5, 5, 5, 7. The 4th value is the median.

\[ \text{Median} = 5 \text{ hours} \]
Solution to question 3 (Mode):

The value 5 appears 3 times — more than any other value.

\[ \text{Mode} = 5 \text{ hours} \]

Sorted: 175, 178, 185, 188, 192, 200. The two middle values are 185 and 188.

\[ \text{Median} = \frac{185 + 188}{2} = \frac{373}{2} = 186.5 \text{ cm} \]
Solution to question 2 (Mean):
\[ \bar{x} = \frac{175+178+185+188+192+200}{6} = \frac{1118}{6} \approx 186.33 \text{ cm} \]

Frequency count: 4 appears 3 times, 9 appears 3 times, 7 appears 2 times, 3 and 11 appear once each.

\[ \text{Modes} = 4 \text{ and } 9 \quad \Rightarrow \text{bimodal} \]
Solution to question 2 (Mode of Set B):

Each value in Set B appears exactly once. No value occurs more frequently than any other.

\[ \text{Set B has no mode.} \]

The highest frequency is 10, corresponding to \(x = 2\).

\[ \text{Mode} = 2 \text{ books} \]

Cumulative frequencies: score 5 → 3; score 6 → 11; score 7 → 25; score 8 → 35; score 9 → 40.

The 20th and 21st values both fall in the score-7 group (cumulative frequency reaches 25 there).

\[ \text{Median} = 7 \]
Solution to question 2 (Mean):
\[ \sum f_i x_i = (5)(3)+(6)(8)+(7)(14)+(8)(10)+(9)(5) = 15+48+98+80+45 = 286 \]
\[ \bar{x} = \frac{286}{40} = 7.15 \]

Re-checking with \(\bar{x} = 3.2\): let us use integer arithmetic — multiplying both sides by 5:

\[ 5(77+3f) = 16(26+f) \implies 385+15f = 416+16f \implies f = -31 \]

Since a negative frequency is impossible, the problem is correctly posed only if we recheck: \(3.2 \times 5 = 16\), so \(16(26+f) = 385+15f \Rightarrow 416+16f = 385+15f \Rightarrow f = -31\). This indicates the stated mean of 3.2 is inconsistent with these fixed frequencies — the corrected mean consistent with \(f = 10\) is:

\[ \frac{77+30}{36} = \frac{107}{36} \approx 2.97 \]

With \(f = 10\), total \(N = 36\) and \(\bar{x} \approx 2.97\). Always verify that a stated mean is achievable before solving.

Solution to question 2:
\[ N = 26 + f = 26 + 10 = 36 \text{ observations} \]

Midpoints: 24.5, 34.5, 44.5, 54.5, 64.5

\[ \sum f_i m_i = (8)(24.5)+(15)(34.5)+(17)(44.5)+(7)(54.5)+(3)(64.5) \]
\[ = 196+517.5+756.5+381.5+193.5 = 2045 \]
\[ \bar{x} \approx \frac{2045}{50} = 40.9 \text{ years} \]
Solution to question 2 (Modal class):

The highest frequency is 17, in the interval 40 – 49.

\[ \text{Modal class} = [40, 49] \]

Cumulative frequencies: 4, 15, 33, 48, 56, 60. The 30th value falls in the class 30 – 39 (cumulative frequency reaches 33 here).

\[ \text{Median class} = [30, 39] \]
Solution to question 2 (Median estimate):
\[ L = 29.5,\quad F = 15,\quad f = 18,\quad h = 10,\quad \frac{N}{2} = 30 \]
\[ \text{Median} \approx 29.5 + \frac{30 – 15}{18} \times 10 = 29.5 + \frac{150}{18} \approx 29.5 + 8.33 = 37.83 \]

The class 400 – 449 has the highest frequency (30).

\[ \text{Modal class} = [400, 449] \]
Solution to question 2 (Mode estimate):
\[ L = 399.5,\quad d_1 = 30 – 22 = 8,\quad d_2 = 30 – 12 = 18,\quad h = 50 \]
\[ \text{Mode} \approx 399.5 + \frac{8}{8+18} \times 50 = 399.5 + \frac{400}{26} \approx 399.5 + 15.38 = 414.88 \]

Sorted: 2800, 2900, 2950, 3000, 3050, 3100, 3200, 18500. Middle pair: 3000 and 3050.

\[ \text{Median} = \frac{3000+3050}{2} = \$3{,}025 \]
Solution to question 2:

The median ($3,025) better represents the typical non-manager income. The mean ($4,937.50) is inflated by the manager’s salary of $18,500, which is an outlier. In skewed distributions, the median is the more robust and representative measure of center.

Sorted (already sorted): 6, 7, 8, 8, 9, 9, 9, 10, 10, 11. Middle pair: 9 and 9.

\[ \text{Median} = \frac{9+9}{2} = 9 \]
\[ \text{Mode} = 9 \text{ (appears 3 times)} \]
Solution to question 2:

The mode (size 9) is the most useful. The owner needs to know which size is sold most frequently to optimize inventory — that is precisely what the mode measures. The mean (8.7) doesn’t correspond to an actual shoe size, and the median (9) coincidentally agrees here but doesn’t directly capture frequency of sale.