Extra Practice Questions
Statistics (R.D. Sharma Level)
1. The mean of the following frequency distribution is 50, but the frequencies f₁ and f₂ in class 20-40 and 60-80 are missing. Find the missing frequencies.
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | Total |
|---|---|---|---|---|---|---|
| Frequency | 17 | f₁ | 32 | f₂ | 19 | 120 |
Step 1: Form the first equation from the sum of frequencies.
17 + f₁ + 32 + f₂ + 19 = 120 => 68 + f₁ + f₂ = 120 => f₁ + f₂ = 52 —(1)
Step 2: Use the Assumed Mean Method to form the second equation. Let A = 50.
| Class | Freq (fᵢ) | xᵢ | dᵢ=xᵢ-A | fᵢdᵢ |
|---|---|---|---|---|
| 0-20 | 17 | 10 | -40 | -680 |
| 20-40 | f₁ | 30 | -20 | -20f₁ |
| 40-60 | 32 | 50 | 0 | 0 |
| 60-80 | f₂ | 70 | 20 | 20f₂ |
| 80-100 | 19 | 90 | 40 | 760 |
| Total | 120 | 80 – 20f₁ + 20f₂ |
Mean = A + (Σfᵢdᵢ / Σfᵢ) => 50 = 50 + (80 – 20f₁ + 20f₂) / 120.
0 = 80 – 20f₁ + 20f₂ => 20f₁ – 20f₂ = 80 => f₁ – f₂ = 4 —(2)
Step 3: Solve the two linear equations.
Adding (1) and (2): 2f₁ = 56 => f₁ = 28.
Substitute f₁ in (1): 28 + f₂ = 52 => f₂ = 24.
Answer: f₁ = 28, f₂ = 24.
2. If the median of the following data is 32.5, find the missing frequencies f₁ and f₂.
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
|---|---|---|---|---|---|---|---|---|
| Frequency | f₁ | 5 | 9 | 12 | f₂ | 3 | 2 | 40 |
Step 1: Create the cumulative frequency table and form the first equation.
| Class Interval | Frequency (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 0-10 | f₁ | f₁ |
| 10-20 | 5 | f₁ + 5 |
| 20-30 | 9 | f₁ + 14 |
| 30-40 | 12 | f₁ + 26 |
| 40-50 | f₂ | f₁ + f₂ + 26 |
| 50-60 | 3 | f₁ + f₂ + 29 |
| 60-70 | 2 | f₁ + f₂ + 31 |
Total frequency is f₁ + f₂ + 31 = 40 => f₁ + f₂ = 9 —(1)
Step 2: Use the median formula.
Median = 32.5, which lies in the class 30-40.
l = 30, N = 40 (N/2=20), cf = f₁+14, f = 12, h = 10.
32.5 = 30 + [ (20 – (f₁+14)) / 12 ] * 10
2.5 = [ (6 – f₁) / 12 ] * 10 => 2.5 * 1.2 = 6 – f₁ => 3 = 6 – f₁ => f₁ = 3.
Substitute into eq (1): 3 + f₂ = 9 => f₂ = 6.
Answer: f₁ = 3, f₂ = 6.
… (8 more questions follow)
3. The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
| Class Interval | 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 |
Step 1: Create the cf table and form the first equation.
2+5+x+12+17+20+y+9+7+4 = 100 => 76 + x + y = 100 => x + y = 24 —(1)
| Class Interval | Frequency (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 0-100 | 2 | 2 |
| 100-200 | 5 | 7 |
| 200-300 | x | 7+x |
| 300-400 | 12 | 19+x |
| 400-500 | 17 | 36+x |
| 500-600 | 20 | 56+x |
| … | … | … |
Step 2: Use the median formula.
Median = 525, which lies in the class 500-600.
l = 500, N=100 (N/2=50), cf = 36+x, f = 20, h = 100.
525 = 500 + [ (50 – (36+x)) / 20 ] * 100
25 = [ (14 – x) / 20 ] * 100 => 25 = (14-x) * 5 => 5 = 14 – x => x = 9.
Substitute into eq (1): 9 + y = 24 => y = 15.
Answer: x = 9, y = 15.
4. Find the mean, mode and median for the following data.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency | 8 | 16 | 36 | 34 | 6 |
| Class | Freq (fᵢ) | cf | xᵢ | dᵢ=xᵢ-25 | fᵢdᵢ |
|---|---|---|---|---|---|
| 0-10 | 8 | 8 | 5 | -20 | -160 |
| 10-20 | 16 | 24 | 15 | -10 | -160 |
| 20-30 | 36 | 60 | 25(A) | 0 | 0 |
| 30-40 | 34 | 94 | 35 | 10 | 340 |
| 40-50 | 6 | 100 | 45 | 20 | 120 |
| Total | N=100 | Σfᵢdᵢ=140 |
Mean: x̄ = 25 + (140/100) = 25 + 1.4 = 26.4.
Median: N/2=50. Median class is 20-30. l=20, cf=24, f=36, h=10.
Median = 20 + [(50-24)/36]*10 = 20 + 260/36 ≈ 20 + 7.22 = 27.22.
Mode: Modal class is 20-30. l=20, f₁=36, f₀=16, f₂=34, h=10.
Mode = 20 + [(36-16)/(2*36-16-34)]*10 = 20 + [20/22]*10 ≈ 20 + 9.09 = 29.09.
5. The mode of the following distribution is 36. Find the missing frequency (x).
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|---|
| Frequency | 8 | 10 | x | 16 | 12 | 6 | 7 |
Mode = 36, which lies in the class 30-40. So, this is the modal class.
l=30, f₁=16, f₀=x, f₂=12, h=10.
Mode = l + [ (f₁ - f₀) / (2f₁ - f₀ - f₂) ] * h
36 = 30 + [ (16 – x) / (2*16 – x – 12) ] * 10
6 = [ (16 – x) / (20 – x) ] * 10
6(20 – x) = 10(16 – x)
120 – 6x = 160 – 10x
4x = 40 => x = 10.
Answer: The missing frequency is 10.
6. Convert the following data to a ‘less than’ type distribution and draw its ogive.
| Class Interval | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 8 | 12 | 24 | 38 | 16 |
‘Less than’ type cumulative frequency distribution:
| Upper Class Limit | Cumulative Frequency | Point for Ogive |
|---|---|---|
| Less than 55 | 2 | (55, 2) |
| Less than 60 | 2 + 8 = 10 | (60, 10) |
| Less than 65 | 10 + 12 = 22 | (65, 22) |
| Less than 70 | 22 + 24 = 46 | (70, 46) |
| Less than 75 | 46 + 38 = 84 | (75, 84) |
| Less than 80 | 84 + 16 = 100 | (80, 100) |
To draw the ogive: Plot the points (55,2), (60,10), (65,22), (70,46), (75,84), (80,100) on a graph paper and join them with a free-hand smooth curve. The x-axis represents the upper class limits, and the y-axis represents the cumulative frequency.
7. If the mean of the distribution is 27, find the value of p.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency | 8 | p | 12 | 13 | 10 |
Using the Direct Method.
| Class | Freq (fᵢ) | xᵢ | fᵢxᵢ |
|---|---|---|---|
| 0-10 | 8 | 5 | 40 |
| 10-20 | p | 15 | 15p |
| 20-30 | 12 | 25 | 300 |
| 30-40 | 13 | 35 | 455 |
| 40-50 | 10 | 45 | 450 |
| Total | 43+p | 1245+15p |
Mean = Σfᵢxᵢ / Σfᵢ => 27 = (1245 + 15p) / (43 + p).
27(43 + p) = 1245 + 15p
1161 + 27p = 1245 + 15p
12p = 84 => p = 7.
Answer: p = 7.
8. The mean of 100 observations is 40. It was found that an observation 53 was misread as 83. Find the correct mean.
Initial Mean = 40, Number of observations (N) = 100.
Incorrect Sum = Mean × N = 40 × 100 = 4000.
Correct Sum = Incorrect Sum – Incorrect Value + Correct Value
= 4000 – 83 + 53 = 3970.
Correct Mean = Correct Sum / N = 3970 / 100 = 39.7.
Answer: The correct mean is 39.7.
9. Find the median of the data, using an empirical relation, when it is given that mode = 12.4 and mean = 10.5.
The empirical relationship between mean, median, and mode is:
Mode = 3 Median – 2 Mean
12.4 = 3 Median – 2(10.5)
12.4 = 3 Median – 21
12.4 + 21 = 3 Median
33.4 = 3 Median
Median = 33.4 / 3 ≈ 11.13.
Answer: The median is approximately 11.13.
10. For the following distribution, calculate the mode.
| Marks | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 | Below 60 |
|---|---|---|---|---|---|---|
| No. of Students | 3 | 12 | 27 | 57 | 75 | 80 |
First, convert the ‘less than’ cumulative frequency data into a normal frequency distribution.
| Class Interval | Frequency (fᵢ) |
|---|---|
| 0-10 | 3 |
| 10-20 | 12 – 3 = 9 |
| 20-30 | 27 – 12 = 15 |
| 30-40 | 57 – 27 = 30 |
| 40-50 | 75 – 57 = 18 |
| 50-60 | 80 – 75 = 5 |
The maximum frequency is 30, so the modal class is 30-40.
l=30, f₁=30, f₀=15, f₂=18, h=10.
Mode = 30 + [ (30 – 15) / (2*30 – 15 – 18) ] * 10
= 30 + [ 15 / (60 – 33) ] * 10 = 30 + 150/27 ≈ 30 + 5.56 = 35.56.
Answer: The mode is approximately 35.56.
