22. Measures of Central Tendency (Mean, Median, Mode) Previous Year Questions

Sum of first 9 natural numbers (1 to 9) = 9(10)/2 = 45. Mean = 45 / 9 = 5 ≠ 4.5. Hence Assertion (A) is false and Reason (R) is true. [1 Mark]

∑f = 24. ∑fx = 30 + 22 + 24 + 78 + 42 + 75 + 48 = 319.
(a) Mean = 319 / 24 ≈ 13.29 ⇒ 13 (nearest whole number). [1.5 Marks]
(b) N = 24 (even) ⇒ Median = (12th + 13th value)/2. Looking at cumulative frequencies: cf for 10=3, 11=5, 12=7, 13=13. Both 12th and 13th values lie in x = 13. Median = 13. [1.5 Marks]

Correct Option: (c) 62
n = 10 ⇒ Median = ½ [x + (x + 4)] = x + 2.
x + 2 = 64 ⇒ x = 62. [1 Mark]

Assumed Mean A = 125, i = 10. Σf = 50, Σfu = −10 − 10 + 0 + 8 + 12 + 12 = 12. [1.5 Marks]
Mean = A + [(Σfu/Σf) × i] = 125 + [(12/50) × 10] = 125 + 2.4 = 127.4 minutes. [1.5 Marks]

Median and Quartiles can be obtained from an Ogive (cumulative frequency curve). Mode is determined from a Histogram. The Mean requires arithmetic computation and cannot be read directly from standard curves. Correct Option: (d) [1 Mark]

Let Assumed Mean $A = 25$ (midpoint of 20–30). Total frequency $\sum f = 3 + 8 + 14 + 9 + 4 + 2 = 40$.1 Mark
Sum of deviations $\sum fd = -60 - 80 + 0 + 90 + 80 + 60 = 90$.1 Mark
Shortcut formula: $\bar{x} = A + \frac{\sum fd}{\sum f} = 25 + \frac{90}{40} = 25 + 2.25 = \mathbf{27.25\text{ marks}}$.1 Mark

Step 1 Calculate $\sum f_i$ and $\sum f_i x_i$:
$\sum f_i = 3 + 7 + f + 9 + 6 = 25 + f$. [1 Mark]
$\sum f_i x_i = (5 \times 3) + (10 \times 7) + (15 \times f) + (20 \times 9) + (25 \times 6)$
$= 15 + 70 + 15f + 180 + 150 = 415 + 15f$. [1 Mark]
Step 2 Applying mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:
$$16 = \frac{415 + 15f}{25 + f} \implies 16(25 + f) = 415 + 15f$$
$$400 + 16f = 415 + 15f \implies 16f - 15f = 415 - 400 \implies \mathbf{f = 15}$$
Answer: $f = 15$. [1 Mark]

Step 1 Prepare Cumulative Frequency ($cf$) Table:
Marks 1: $f=1, cf=1$
Marks 2: $f=2, cf=3$
Marks 3: $f=3, cf=6$
Marks 4: $f=3, cf=9$
Marks 5: $f=6, cf=15$
Marks 6: $f=10, cf=25$
Marks 7: $f=5, cf=30$
Marks 8: $f=4, cf=34$
Marks 9: $f=3, cf=37$
Marks 10: $f=3, cf=40 \implies N = 40$ (even). [1 Mark]
Step 2 (i) Calculate Median:
$\text{Median} = \frac{1}{2} \left[ \left(\frac{N}{2}\right)\text{th term} + \left(\frac{N}{2} + 1\right)\text{th term} \right] = \frac{20\text{th term} + 21\text{st term}}{2}$.
Looking at $cf$, both 20th and 21st terms fall in the group where $cf = 25$, corresponding to Marks = 6.
$\therefore \text{Median} = \frac{6 + 6}{2} = \mathbf{6\text{ marks}}$. [2 Marks]
Step 3 (ii) Calculate Mode:
Mode is the observation having the maximum frequency.
Here, maximum frequency is $f = 10$, which occurs for Marks = 6.
$\therefore \mathbf{\text{Mode} = 6\text{ marks}}$. [1 Mark]

Step 1 Prepare Assumed Mean Table with $A = 45$:
[1.5 Marks]
Step 2 Apply Assumed Mean Formula:
$\text{Mean } \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} = 45 + \frac{-140}{30} = 45 - 4.6667 = \mathbf{40.33}$.
Average number of patients = 40.33. [1.5 Marks]

Solution If every observation $x_i$ is increased by a constant $k$, the arithmetic mean is also increased by $k$. Here $k = 2$, so new mean $= \mathbf{m + 2}$.
Correct Answer: (c) m + 2 [1 Mark]

Step 1 Prepare step-deviation table with Assumed Mean $A = 32.5$, class width $h = 5$:
$u_i = \frac{x_i - 32.5}{5}$:
$20-25: x_1 = 22.5, u_1 = -2, f_1 = 10, f_1 u_1 = -20$
$25-30: x_2 = 27.5, u_2 = -1, f_2 = 7, f_2 u_2 = -7$
$30-35: x_3 = 32.5, u_3 = 0, f_3 = 9, f_3 u_3 = 0$
$35-40: x_4 = 37.5, u_4 = +1, f_4 = 13, f_4 u_4 = +13$
$40-45: x_5 = 42.5, u_5 = +2, f_5 = 5, f_5 u_5 = +10$
$45-50: x_6 = 47.5, u_6 = +3, f_6 = 6, f_6 u_6 = +18$
$\sum f_i = 10 + 7 + 9 + 13 + 5 + 6 = 50$.
$\sum f_i u_i = -27 + 41 = 14$. [1.5 Marks]
Step 2 Apply Step-Deviation Formula:
$\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 32.5 + \left(\frac{14}{50}\right) \times 5 = 32.5 + 1.4 = \mathbf{33.9}$.
Rounding to nearest whole number: $\mathbf{\bar{x} \approx 34\text{ marks}}$. [1.5 Marks]

Step 1 Find midpoints ($x_i$) and products ($f_i x_i$):
• 0 - 10: $x_1 = 5, f_1 = 7 \implies f_1 x_1 = 35$
• 10 - 20: $x_2 = 15, f_2 = a \implies f_2 x_2 = 15a$
• 20 - 30: $x_3 = 25, f_3 = 8 \implies f_3 x_3 = 200$
• 30 - 40: $x_4 = 35, f_4 = 10 \implies f_4 x_4 = 350$
• 40 - 50: $x_5 = 45, f_5 = 5 \implies f_5 x_5 = 225$
$$\sum f_i = 30 + a, \quad \sum f_i x_i = 810 + 15a$$ [1 Mark]
Step 2 Apply Mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 24$:
$$24 = \frac{810 + 15a}{30 + a} \implies 24(30 + a) = 810 + 15a$$ [1 Mark]
$$720 + 24a = 810 + 15a \implies 9a = 90 \implies \mathbf{a = 10}$$ [1 Mark]
Answer: $a = 10$.

Step 1: Find x For $n = 10$ (even number of terms):
$$\text{Median} = \frac{\left(\frac{10}{2}\right)\text{th term} + \left(\frac{10}{2} + 1\right)\text{th term}}{2} = \frac{5\text{th term} + 6\text{th term}}{2}$$ $$48 = \frac{x + (x + 4)}{2} = \frac{2x + 4}{2} = x + 2$$ $$x = 48 - 2 = \mathbf{46}$$ [2 Marks]
Step 2: Find Mode Substitute $x = 46$ into the dataset:
$$13, 35, 43, \mathbf{46}, \mathbf{46}, 50, 55, 61, 71, 80$$ $46$ appears with the highest frequency (twice).
$$\mathbf{\text{Mode} = 46}$$ [1 Mark]
Answers: $x = 46$; Mode $= 46$.

Step 1 Let Assumed Mean $A = 35$, class width $h = 10$, $u_i = \frac{x_i - 35}{10}$.
• 0 - 10: $x_1 = 5, f_1 = 10, u_1 = -3 \implies f_1 u_1 = -30$
• 10 - 20: $x_2 = 15, f_2 = 9, u_2 = -2 \implies f_2 u_2 = -18$
• 20 - 30: $x_3 = 25, f_3 = 25, u_3 = -1 \implies f_3 u_3 = -25$
• 30 - 40: $x_4 = 35, f_4 = 30, u_4 = 0 \implies f_4 u_4 = 0$
• 40 - 50: $x_5 = 45, f_5 = 16, u_5 = 1 \implies f_5 u_5 = 16$
• 50 - 60: $x_6 = 55, f_6 = 10, u_6 = 2 \implies f_6 u_6 = 20$
$$\sum f_i = 100, \quad \sum f_i u_i = -37$$ [2 Marks]
Step 2 Apply Step-Deviation formula:
$$\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 35 + \left(\frac{-37}{100}\right) \times 10 = 35 - 3.7 = \mathbf{31.3}$$ [2 Marks]
Answer: $\text{Mean} = 31.3$.