Question 1: The mean of the following frequency distribution is 53. But the frequencies $f_1$ and $f_2$ in the classes 20-40 and 60-80 are missing. Find the missing frequencies if the total frequency is 100.
Class 0-20 20-40 40-60 60-80 80-100 Frequency 15 $f_1$ 21 $f_2$ 17
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Solution: Step 1: Sum of frequencies: $15 + f_1 + 21 + f_2 + 17 = 100 \Rightarrow f_1 + f_2 = 47$.Step 2: Mean formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. $x_i$: 10, 30, 50, 70, 90.Step 3: $\sum f_i x_i = 150 + 30f_1 + 1050 + 70f_2 + 1530 = 2730 + 30f_1 + 70f_2$.Step 4: $53 = \frac{2730 + 30f_1 + 70f_2}{100} \Rightarrow 5300 = 2730 + 30f_1 + 70f_2 \Rightarrow 3f_1 + 7f_2 = 257$.Step 5: Solving $f_1+f_2=47$ and $3f_1+7f_2=257$: Multiply eq1 by 3: $3f_1+3f_2=141$. Subtract: $4f_2 = 116 \Rightarrow f_2 = 29$. $f_1 = 18$.
Question 2: The median of the following data is 525. Find the values of $x$ and $y$, if the total frequency is 100.
Class 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900-1000 Frequency 2 5 $x$ 12 17 20 $y$ 9 7 4
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Solution: Step 1: $76 + x + y = 100 \Rightarrow x + y = 24$.Step 2: Median is 525, so median class is 500-600. $l=500, h=100, f=20, cf=36+x$.Step 3: $525 = 500 + (\frac{50 - (36+x)}{20}) \times 100 \Rightarrow 25 = (14-x) \times 5 \Rightarrow 5 = 14-x \Rightarrow x=9$.Step 4: $y = 24 - 9 = 15$.Answer: $x=9, y=15$.
Question 3: If the mode of the following distribution is 55, find the value of $x$.
Class 0-15 15-30 30-45 45-60 60-75 75-90 Frequency 6 7 $x$ 15 10 8
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Solution: Step 1: Mode is 55, so modal class is 45-60. $l=45, f_1=15, f_0=x, f_2=10, h=15$.Step 2: $55 = 45 + (\frac{15-x}{2(15)-x-10}) \times 15 \Rightarrow 10 = \frac{15-x}{20-x} \times 15$.Step 3: Divide by 5: $2 = \frac{15-x}{20-x} \times 3 \Rightarrow 2(20-x) = 3(15-x) \Rightarrow 40-2x = 45-3x \Rightarrow x=5$.
Question 4: Calculate the median of the following data:
Marks < 10 < 20 < 30 < 40 < 50 Students 5 17 31 41 49
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Solution: Step 1: Convert to frequency distribution. Classes: 0-10, 10-20, 20-30, 30-40, 40-50. Frequencies: 5, 12, 14, 10, 8.Step 2: $N=49, N/2=24.5$. Median class 20-30 ($cf=17$).Step 3: Median $= 20 + (\frac{24.5-17}{14}) \times 10 = 20 + \frac{7.5}{1.4} = 20 + 5.36 = 25.36$.
Question 5: The following table gives the production yield per hectare of wheat of 100 farms of a village. Change the distribution to a 'more than type' distribution and find its mean.
Yield 50-55 55-60 60-65 65-70 70-75 75-80 Farms 2 8 12 24 38 16
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Solution: Step 1: More than type: >50 (100), >55 (98), >60 (90), >65 (78), >70 (54), >75 (16).Step 2: To find mean, use original classes. $x_i$: 52.5, 57.5, 62.5, 67.5, 72.5, 77.5.Step 3: Assumed mean $a=67.5$. $d_i$: -15, -10, -5, 0, 5, 10. $u_i$: -3, -2, -1, 0, 1, 2.Step 4: $\sum f_i u_i = -6 -16 -12 + 0 + 38 + 32 = 36$.Step 5: Mean $= 67.5 + \frac{36}{100} \times 5 = 67.5 + 1.8 = 69.3$.
Question 6: Find the median of the following data:
Class 10-19 20-29 30-39 40-49 Frequency 5 8 12 5
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Solution: Step 1: Convert to exclusive classes: 9.5-19.5, 19.5-29.5, 29.5-39.5, 39.5-49.5.Step 2: $N=30, N/2=15$. CF: 5, 13, 25, 30. Median class 29.5-39.5.Step 3: Median $= 29.5 + (\frac{15-13}{12}) \times 10 = 29.5 + \frac{20}{12} = 29.5 + 1.67 = 31.17$.
Question 7: If the mode of a distribution exceeds its mean by 24, then by how much does the mode exceed the median?
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Solution: Step 1: Given Mode - Mean = 24.Step 2: Empirical relation: Mode = 3 Median - 2 Mean.Step 3: Mode - Mean = 3 Median - 3 Mean = 3(Median - Mean).Step 4: $24 = 3(\text{Median} - \text{Mean}) \Rightarrow \text{Median} - \text{Mean} = 8$.Step 5: Mode - Median = (Mode - Mean) - (Median - Mean) = $24 - 8 = 16$.
Question 8: Find the unknown entries $a, b, c, d, e, f$ in the following distribution of heights of students in a class:
Height 150-155 155-160 160-165 165-170 170-175 175-180 Freq 12 $b$ 10 $d$ $e$ 2 Cum Freq $a$ 25 $c$ 43 48 $f$
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Solution: Step 1: $a = 12$.Step 2: $12 + b = 25 \Rightarrow b = 13$.Step 3: $c = 25 + 10 = 35$.Step 4: $35 + d = 43 \Rightarrow d = 8$.Step 5: $43 + e = 48 \Rightarrow e = 5$.Step 6: $f = 48 + 2 = 50$.
Question 9: If the mean of a set of observations $x_1, x_2, \dots, x_n$ is $\bar{x}$, then find the mean of $ax_1+b, ax_2+b, \dots, ax_n+b$.
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Solution: Step 1: New Mean $= \frac{\sum (ax_i + b)}{n} = \frac{a\sum x_i + \sum b}{n}$.Step 2: $= \frac{a(n\bar{x}) + nb}{n} = \frac{n(a\bar{x} + b)}{n}$.Step 3: New Mean $= a\bar{x} + b$.
Question 10: Find the mean of the following data using the Step Deviation Method:
Class 0-50 50-100 100-150 150-200 200-250 250-300 Frequency 17 35 43 40 21 24
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Solution: Step 1: $x_i$: 25, 75, 125, 175, 225, 275. Let $a=125, h=50$.Step 2: $u_i$: -2, -1, 0, 1, 2, 3.Step 3: $f_i u_i$: -34, -35, 0, 40, 42, 72. Sum $= 85$. Total Freq $= 180$.Step 4: Mean $= 125 + \frac{85}{180} \times 50 = 125 + 23.61 = 148.61$.