In This Post we are providing Chapter 14 Statistics NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Statistics Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Statistics NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
| Number of Plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |
| Number of Houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Answer
| No. of plants (Class interval) | No. of houses (fi) | Mid-point (xi) | fixi |
| 0-2 | 1 | 1 | 1 |
| 2-4 | 2 | 3 | 6 |
| 4-6 | 1 | 5 | 5 |
| 6-8 | 5 | 7 | 35 |
| 8-10 | 6 | 9 | 54 |
| 10-12 | 2 | 11 | 22 |
| 12-14 | 3 | 13 | 39 |
| Sum fi = 20 | Sum fixi = 162 |
Mean = x̄ = ∑fixi /∑fi = 162/20 = 8.1
We would use direct method because the numerical value of fi and xi are small.
2. Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in Rs.) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Answer
Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui = (xi – A)/h = ui = (xi – 150)/20
| Daily wages (Class interval) | Number of workers frequency (fi) | Mid-point (xi) | ui = (xi – 150)/20 | fiui |
| 100-120 | 12 | 110 | -2 | -24 |
| 120-140 | 14 | 130 | -1 | -14 |
| 140-160 | 8 | 150 | 0 | 0 |
| 160-180 | 6 | 170 | 1 | 6 |
| 180-200 | 10 | 190 | 2 | 20 |
| Total | Sum fi = 50 | Sum fiui = -12 |
Thus, mean daily wage = Rs. 145.20
3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Answer
Here, the value of mid-point (xi) mean x̄ = 18
| Class interval | Number of children (fi) | Mid-point (xi) | fixi |
| 11-13 | 7 | 12 | 84 |
| 13-15 | 6 | 14 | 84 |
| 15-17 | 9 | 16 | 144 |
| 17-19 | 13 | 18 = A | 234 |
| 19-21 | f | 20 | 20f |
| 21-23 | 5 | 22 | 110 |
| 23-25 | 4 | 24 | 96 |
| Total | fi = 44+f | Sum fixi = 752+20f |
Mean = x̄ = ∑fixi /∑fi = (752+20f)/(44+f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 – 752 = 20f – 18f
⇒ 40 = 2f
⇒ f = 20
Page No: 271
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Answer
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5
| Class Interval | Number of women (fi) | Mid-point (xi) | ui = (xi – 75.5)/h | fiui |
| 65-68 | 2 | 66.5 | -3 | -6 |
| 68-71 | 4 | 69.5 | -2 | -8 |
| 71-74 | 3 | 72.5 | -1 | -3 |
| 74-77 | 8 | 75.5 | 0 | 0 |
| 77-80 | 7 | 78.5 | 1 | 7 |
| 80-83 | 4 | 81.5 | 3 | 8 |
| 83-86 | 2 | 84.5 | 3 | 6 |
| Sum fi= 30 | Sum fiui = 4 |
Mean = x̄ = A + h∑fiui /∑fi = 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Answer
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
| Class Interval | Number of boxes (fi) | Mid-point (xi) | di = xi – A | fidi |
| 49.5-52.5 | 15 | 51 | -6 | 90 |
| 52.5-55.5 | 110 | 54 | -3 | -330 |
| 55.5-58.5 | 135 | 57 = A | 0 | 0 |
| 58.5-61.5 | 115 | 60 | 3 | 345 |
| 61.5-64.5 | 25 | 63 | 6 | 150 |
| Sum fi = 400 | Sum fidi = 75 |
Mean = x̄ = A + ∑fidi /∑fi = 57 + (75/400) = 57 + 0.1875 = 57.19
6. The table below shows the daily expenditure on food of 25 households in a locality.
Here, assumed mean (A) = 225
| Class Interval | Number of households (fi) | Mid-point (xi) | di = xi – A | fidi |
| 100-150 | 4 | 125 | -100 | -400 |
| 150-200 | 5 | 175 | -50 | -250 |
| 200-250 | 12 | 225 | 0 | 0 |
| 250-300 | 2 | 275 | 50 | 100 |
| 300-350 | 2 | 325 | 100 | 200 |
| Sum fi = 25 | Sum fidi = -350 |
Mean = x̄ = A + ∑fidi /∑fi = 225 + (-350/25) = 225 – 14 = 211
The mean daily expenditure on food is 211
7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Answer
| Concentration of SO2 (in ppm) | Frequency (fi) | Mid-point (xi) | fixi |
| 0.00-0.04 | 4 | 0.02 | 0.08 |
| 0.04-0.08 | 9 | 0.06 | 0.54 |
| 0.08-0.12 | 9 | 0.10 | 0.90 |
| 0.12-0.16 | 2 | 0.14 | 0.28 |
| 0.16-0.20 | 4 | 0.18 | 0.72 |
| 0.20-0.24 | 2 | 0.20 | 0.40 |
| Total | Sum fi = 30 | Sum (fixi) = 2.96 |
Mean = x̄ = ∑fixi /∑fi
= 2.96/30 = 0.099 ppm
Page No. 272
8. A class teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.
Number of days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
|---|---|---|---|---|---|---|---|
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Answer
| Class interval | Frequency (fi) | Mid-point (xi) | fixi |
| 0-6 | 11 | 3 | 33 |
| 6-10 | 10 | 8 | 80 |
| 10-14 | 7 | 12 | 84 |
| 14-20 | 4 | 17 | 68 |
| 20-28 | 4 | 24 | 96 |
| 28-38 | 3 | 33 | 99 |
| 38-40 | 1 | 39 | 39 |
| Sum fi = 40 | Sum fixi = 499 |
Mean = x̄ = ∑fixi /∑fi
= 499/40 = 12.48 days
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
| Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-98 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Answer
| Class Interval | Frequency (fi) | (xi) | di = xi – a | ui = di/h | fiui |
| 45-55 | 3 | 50 | -20 | -2 | -6 |
| 55-65 | 10 | 60 | -10 | -1 | -10 |
| 65-75 | 11 | 70 | 0 | 0 | 0 |
| 75-85 | 8 | 80 | 10 | 1 | 8 |
| 85-95 | 3 | 90 | 20 | 2 | 6 |
| Sum fi = 35 | Sum fiui = -2 |
Mean = x̄ = a + (∑fiui /∑fi) х h
= 70 + (-2/35) х 10 = 69.42
Exercise 14.2
1. The following table shows the ages of the patients admitted in a hospital during 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 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer
Modal class = 35 – 45, l = 35, class width (h) = 10, fm = 23, f1 = 21 and f2 = 14
| 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 |
Mean = x̄ = ∑fixi /∑fi
= 2830/80 = 35.37 yr
2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components :
| Lifetime (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Answer
Modal class of the given data is 60–80.
Modal class = 60-80, l = 60, fm = 61, f1 = 52, f2 = 38 and h = 20
3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. 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 |
Answer
Modal class = 1500-2000, l = 1500, fm = 40, f1 = 24, f2 = 33 and h = 500
| 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 |
Mean = x̄ = a + (∑fiui /∑fi) х h
= 2750 + (35/200) х 500
= 2750 – 87.50 = 2662.50
Question 4.
The following distribution gives the state-wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Solution:
Question 5.
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.
Solution:
Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
Solution:
Question 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.
Solution:
Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
Solution:
Question 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 years.
Solution:
Question 4.
The lengths of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table:
Find the median length of the leaves.
Solution:
Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
Find the median lifetime of a lamp.
Solution:
Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:
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.
Solution:
Ex 14.3 Class 10 Maths Question 7.
The distribution below gives the weight of 30 students of a class. Find the median weight of the students.
Solution:
Question 1.
The following distribution gives the daily income of 50 workers of a factory.
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
Ex 14.4 Class 10 Maths Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
Table of Contents
Important Links
Statistics – Quick Revision Notes
Statistics – Most Important Questions
Statistics – Important MCQs
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