Exercise 13.3 Solutions
Statistics
1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
First, we create a table with cumulative frequency to find the median, and other columns for mean and mode.
| Class Interval | Frequency (fᵢ) | Cumulative Freq. (cf) | Class mark (xᵢ) | uᵢ=(xᵢ-135)/20 | fᵢuᵢ |
|---|---|---|---|---|---|
| 65 – 85 | 4 | 4 | 75 | -3 | -12 |
| 85 – 105 | 5 | 9 | 95 | -2 | -10 |
| 105 – 125 | 13 | 22 | 115 | -1 | -13 |
| 125 – 145 | 20 | 42 | 135 (A) | 0 | 0 |
| 145 – 165 | 14 | 56 | 155 | 1 | 14 |
| 165 – 185 | 8 | 64 | 175 | 2 | 16 |
| 185 – 205 | 4 | 68 | 195 | 3 | 12 |
| Total | N = 68 | Σfᵢuᵢ = 7 |
Calculation of Median
N = 68, so N/2 = 34. The cumulative frequency just greater than 34 is 42, so the median class is 125 – 145.
l = 125, N/2 = 34, cf (of preceding class) = 22, f = 20, h = 20.
Median = l + [ (N/2 - cf) / f ] * h = 125 + [ (34 – 22) / 20 ] * 20 = 125 + 12 = 137.
Calculation of Mode
The maximum frequency is 20, so the modal class is 125 – 145.
l = 125, h = 20, f₁ = 20, f₀ = 13, f₂ = 14.
Mode = 125 + [ (20-13) / (2*20 – 13 – 14) ] * 20 = 125 + [ 7 / (40 – 27) ] * 20 = 125 + 140/13 ≈ 125 + 10.77 = 135.77.
Calculation of Mean
Using the table above, A = 135, h = 20, Σfᵢ = 68, Σfᵢuᵢ = 7.
Mean = A + ( (Σfᵢuᵢ / Σfᵢ) * h ) = 135 + ( (7 / 68) * 20 ) = 135 + 140/68 ≈ 135 + 2.05 = 137.05.
Comparison
The three measures of central tendency are close to each other: Mean = 137.05, Median = 137, Mode = 135.77. This indicates a fairly symmetrical distribution of data.
2. If the median of the distribution given below is 28.5, find the values of x and y.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
|---|---|---|---|---|---|---|---|
| Frequency | 5 | x | 20 | 15 | y | 5 | 60 |
Step 1: Create the cumulative frequency table and form the first equation.
| Class Interval | Frequency (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | x | 5 + x |
| 20-30 | 20 | 25 + x |
| 30-40 | 15 | 40 + x |
| 40-50 | y | 40 + x + y |
| 50-60 | 5 | 45 + x + y |
From the table, the total frequency is 45 + x + y. We are given the total is 60.
So, 45 + x + y = 60 => x + y = 15 —(1)
Step 2: Use the median formula to find x.
Given Median = 28.5, which lies in the class interval 20 – 30.
l = 20, N = 60 (so N/2 = 30), cf (of preceding class) = 5+x, f = 20, h = 10.
Median = l + [ (N/2 - cf) / f ] * h
28.5 = 20 + [ (30 – (5+x)) / 20 ] * 10
8.5 = [ (25 – x) / 20 ] * 10
8.5 = (25 – x) / 2
17 = 25 – x => x = 8.
Step 3: Find y.
Substitute x=8 into equation (1): 8 + y = 15 => y = 7.
Answer: x = 8 and y = 7.
3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
First, we convert the ‘below’ type data into a class interval frequency distribution.
| Age (in years) | Number of policy holders (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 15 – 20 | 2 | 2 |
| 20 – 25 | 6 – 2 = 4 | 6 |
| 25 – 30 | 24 – 6 = 18 | 24 |
| 30 – 35 | 45 – 24 = 21 | 45 |
| 35 – 40 | 78 – 45 = 33 | 78 |
| 40 – 45 | 89 – 78 = 11 | 89 |
| 45 – 50 | 92 – 89 = 3 | 92 |
| 50 – 55 | 98 – 92 = 6 | 98 |
| 55 – 60 | 100 – 98 = 2 | 100 |
| Total | N = 100 |
N = 100, so N/2 = 50. The cf just greater than 50 is 78. So, the median class is 35 – 40.
l = 35, N/2 = 50, cf = 45, f = 33, h = 5.
Median = 35 + [ (50 – 45) / 33 ] * 5 = 35 + (5/33) * 5 = 35 + 25/33 ≈ 35 + 0.76 = 35.76 years.
Answer: The median age is 35.76 years.
4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: Find the median length of the leaves.
The class intervals are discontinuous. We make them continuous by subtracting 0.5 from lower limits and adding 0.5 to upper limits.
| Length (in mm) | Number of leaves (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 117.5 – 126.5 | 3 | 3 |
| 126.5 – 135.5 | 5 | 8 |
| 135.5 – 144.5 | 9 | 17 |
| 144.5 – 153.5 | 12 | 29 |
| 153.5 – 162.5 | 5 | 34 |
| 162.5 – 171.5 | 4 | 38 |
| 171.5 – 180.5 | 2 | 40 |
N = 40, so N/2 = 20. The cf just greater than 20 is 29. So, the median class is 144.5 – 153.5.
l = 144.5, N/2 = 20, cf = 17, f = 12, h = 9.
Median = 144.5 + [ (20 – 17) / 12 ] * 9 = 144.5 + (3/12) * 9 = 144.5 + 2.25 = 146.75 mm.
Answer: The median length of the leaves is 146.75 mm.
5. The following table gives the distribution of the life time of 400 neon lamps: Find the median life time of a lamp.
| Life time (in hours) | Number of lamps (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 1500 – 2000 | 14 | 14 |
| 2000 – 2500 | 56 | 70 |
| 2500 – 3000 | 60 | 130 |
| 3000 – 3500 | 86 | 216 |
| 3500 – 4000 | 74 | 290 |
| 4000 – 4500 | 62 | 352 |
| 4500 – 5000 | 48 | 400 |
N = 400, so N/2 = 200. The cf just greater than 200 is 216. So, the median class is 3000 – 3500.
l = 3000, N/2 = 200, cf = 130, f = 86, h = 500.
Median = 3000 + [ (200 – 130) / 86 ] * 500 = 3000 + (70/86) * 500 = 3000 + 35000/86 ≈ 3000 + 406.98 = 3406.98 hours.
Answer: The median life time is 3406.98 hours.
6. 100 surnames were randomly picked up from a local telephone directory… Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Calculation Table:
| Class Interval | Freq. (fᵢ) | cf | xᵢ | uᵢ=(xᵢ-8.5)/3 | fᵢuᵢ |
|---|---|---|---|---|---|
| 1-4 | 6 | 6 | 2.5 | -2 | -12 |
| 4-7 | 30 | 36 | 5.5 | -1 | -30 |
| 7-10 | 40 | 76 | 8.5 (A) | 0 | 0 |
| 10-13 | 16 | 92 | 11.5 | 1 | 16 |
| 13-16 | 4 | 96 | 14.5 | 2 | 8 |
| 16-19 | 4 | 100 | 17.5 | 3 | 12 |
| Total | N=100 | Σfᵢuᵢ = -6 |
Median
N=100, N/2=50. Median class is 7 – 10.
l=7, cf=36, f=40, h=3.
Median = 7 + [ (50-36)/40 ] * 3 = 7 + (14*3)/40 = 7 + 1.05 = 8.05.
Mean
Mean = A + (Σfᵢuᵢ/Σfᵢ)*h = 8.5 + (-6/100)*3 = 8.5 – 0.18 = 8.32.
Mode
Modal class is 7 – 10.
l=7, f₁=40, f₀=30, f₂=16, h=3.
Mode = 7 + [(40-30)/(2*40-30-16)]*3 = 7 + [10/34]*3 = 7 + 0.88 = 7.88.
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kg) | Number of students (fᵢ) | Cumulative Freq. (cf) |
|---|---|---|
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 5 |
| 50 – 55 | 8 | 13 |
| 55 – 60 | 6 | 19 |
| 60 – 65 | 6 | 25 |
| 65 – 70 | 3 | 28 |
| 70 – 75 | 2 | 30 |
N = 30, so N/2 = 15. The cf just greater than 15 is 19.
So, the median class is 55 – 60.
Wait, the cf just greater than 15 is 19, which corresponds to class 55-60. The cf of the *preceding* class is 13.
l = 55, N/2 = 15, cf = 13, f = 6, h = 5.
Median = 55 + [ (15 – 13) / 6 ] * 5 = 55 + (2/6) * 5 = 55 + 5/3 ≈ 55 + 1.67 = 56.67 kg.
Answer: The median weight is 56.67 kg.
