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Statistics is a branch of mathematics that deals with data collection, organisation, analysis, interpretation, and presentation. Statistics is extremely useful in real-life situations since it is simple to comprehend when we express data in a single number that reflects all values. The measure of central tendency is the name given to this number.
Exercise 14.2 solutions:
Q1. The following table illustrates the ages of hospital patients over the course of a year:
| Age (in years) | 5 – 15 | 15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 |
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
From the data given above, find the mean and mode. Compare and interpret the two measures of central tendency.
Solution: Consider the class interval with the highest frequency to determine the modal class.
Because the highest frequency is 23, the modal class is 35 – 45.
I = 35
h = class width = 10
fm = 23
f₁ = 21 and f₂ = 14
The formula for determining the mode is as follows:
Mode = I + [(fm- f₁)/(2fm-f₁-f₂)] x h
When the values in the formula are substituted, we get
| Class interval | Frequency (fi) | Mid – point (xi) | fixi |
| 5 – 15 | 6 | 10 | 60 |
| 15 – 25 | 11 | 20 | 220 |
| 25 – 35 | 21 | 30 | 630 |
| 35 – 45 | 23 | 40 | 920 |
| 45 – 55 | 14 | 50 | 700 |
| 55 – 65 | 5 | 60 | 300 |
| Sum fi = 80 | Sum fixi = 2830 |
Mode = 35 + [(23-21)/(46-21-14)] x 10
Mode = 35 + (20/11) = 35+1.8
Mode = 36.8 year
As a result, the data’s mode is 36.8 years.
Calculation of the mean:
To begin, use the formula to get the midpoint,
xᵢ = (upper limit + lower limit)/2
The formula for mean is
Mean = X̄ = ∑fᵢxᵢ /∑fᵢ
= 2830/80
= 35.37 years
As a result, the mean of the data is calculated as 35.37 years.
Q2. The information on the observed lives (in hours) of 225 electrical components is provided in the following table:
| Lifetime (in hours) | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the component’s modal lifespan.
Solution: The modal class based on the data is 60–80.
I = 60
The frequencies are:
fm = 61, f1 = 52, f2 = 38, and h = 20
The formula to calculate the mode,
Mode = l+ [(fm-f₁)/(2fm-f₁-f₂)]×h
When the values in the formula are substituted, we get
Mode = 60 + [(61-52)/(122-52-38)]
Mode = 60 + [(9×20)/32]
Mode = 60 + (45/8) = 60 + 5.625
As a result, the components’ modal lifetime is 65.625 hours.
Q3. The distribution of total monthly household expenditure for 200 families in a hamlet is shown in the graph below. Determine the families’ modal monthly expenditure. Also, find the mean monthly expenditure:
| Expenditure | Number of families |
| 1000 – 1500 | 24 |
| 1500 – 2000 | 40 |
| 2000 – 2500 | 33 |
| 2500 – 3000 | 28 |
| 3000 – 3500 | 30 |
| 3500 – 4000 | 22 |
| 4000 – 4500 | 16 |
| 4500 – 5000 | 7 |
Solution: Given,
Modal class = 1500-2000
I = 1500
The frequencies are:
fm = 40, f₁ = 24, f₂ = 33 and
h = 500
The formula for mode,
Mode = l+ [(fm-f₁)/(2fm-f₁-f₂)]×h
When the values in the formula are substituted, we get,
Mode = 1500 + [(40-24)/(80-24-33)] x 500
Mode = 1500 + [(16×500)/23]
Mode = 1500 + (8000/23) = 1500 + 347.83
As a result, the modal monthly spending of the families is 1847.83 rupees.
Calculation for the mean:
To begin, use the formula to get the midpoint,
xᵢ = (upper limit + lower limit)/2
Suppose a mean, A = 2750
| Class interval | fi | xi | di = xi – a | ui = di/h | fiui |
| 1000 – 1500 | 24 | 1250 | -1500 | -3 | -72 |
| 1500 – 2000 | 40 | 1750 | -1000 | -2 | -80 |
| 2000 – 2500 | 33 | 2250 | -500 | -1 | -33 |
| 2500 – 3000 | 28 | 2750 | 0 | 0 | 0 |
| 3000 – 3500 | 30 | 3250 | 500 | 1 | 30 |
| 3500 – 4000 | 22 | 3750 | 1000 | 2 | 44 |
| 4000 – 4500 | 16 | 4250 | 1500 | 3 | 48 |
| 4500 – 5000 | 7 | 4750 | 2000 | 4 | 28 |
| fi = 200 | fiui = -35 |
The following is the formula for determining the mean:
Mean = X̄ = a + (∑fᵢuᵢ /∑fᵢ) x h
Fill in the blanks in the provided formula
= 2750 + (-35/200) x 500
= 2750 + 87.50
= 2662.50
As a result, the average monthly expenditure of the families is Rs. 2662.50.
Q4. The following table shows the teacher-to-student ratio in India’s higher secondary schools by state. Determine the data’s mode and mean. Interpret the two measures-
| Number of students per teacher | Number of states/ U.T. |
| 15 – 20 | 3 |
| 20 – 25 | 8 |
| 25 – 30 | 9 |
| 30 – 35 | 10 |
| 35 – 40 | 3 |
| 40 – 45 | 0 |
| 45 – 50 | 0 |
| 50 – 55 | 2 |
Solution: Given data,
Modal class = 30-35
I = 30
h = class width = 5
fm = 10, f₁ = 9 and f₂ = 3
Mode Formula:
Mode = l+ [(fm-f₁)/(2fm-f₁-f₂)]×h
When the values in the formula are substituted, we get,
Mode = 30+((10-9)/(20-9-3))×5
Mode = 30+(5/8) = 30+0.625
Mode = 30.625
As a result, the given data’s mode is 30.625.
Calculation for the mean:
To begin, use the formula to get the midpoint,
xᵢ = (upper limit + lower limit)/2
| Class interval | Frequency (fi) | Mid-point (xi) | fixi |
| 15 – 20 | 3 | 17.5 | 52.5 |
| 20 – 25 | 8 | 22.5 | 180.0 |
| 25 – 30 | 9 | 27.5 | 247.5 |
| 30 – 35 | 10 | 32.5 | 325.0 |
| 35 – 40 | 3 | 37.5 | 112.5 |
| 40 – 45 | 0 | 42.5 | 0 |
| 45 – 50 | 0 | 47.5 | 0 |
| 50 – 55 | 2 | 52.5 | 105.5 |
| Sum fi = 35 | Sum fixi = 1022.5 |
Mean = x̄ = ∑fᵢxᵢ /∑fᵢ
= 1022.5/35
= 29.2
Therefore the mean becomes 29.2
Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.
Find the mode of the data.
| Run scored | Number of batsman |
| 3000 – 4000 | 4 |
| 4000 – 5000 | 18 |
| 5000 – 6000 | 9 |
| 6000 – 7000 | 7 |
| 7000 – 8000 | 6 |
| 8000 – 9000 | 3 |
| 9000 – 10000 | 1 |
| 10000 – 11000 | 1 |
Solution: The given data is as follows,
Modal class = 4000 – 5000
l = 4000
h = class width = 1000
fm = 18, f1 = 4 and f2 = 9
The formula for mode is
Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h
By substituting the values we get,
Mode = 4000+((18-4)/(36-4-9))×1000
Mode = 4000+(14000/23) = 4000+608.695
Mode = 4608.695
Mode = 4608.7 (approximately)
As a result, the data’s mode is 4608.7 runs.
Conclusion:
The term “statistics” refers to a set of mathematical equations that we use to analyse data. It keeps us informed about what is going on in the globe. Because we live in an information age, statistics are essential.
