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NCERT Solutions

Statistics

Madhya Pradesh Board · Class 10 · Mathematics

NCERT Solutions for Statistics — Madhya Pradesh Board Class 10 Mathematics.

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33 Questions Solved · 3 Sections

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EXERCISE 13.1

1A 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.Show solution
Using the direct method, take class marks xi=1,3,5,7,9,11,13x_i = 1, 3, 5, 7, 9, 11, 13 for the intervals $0\!-
2, 2\!-
4, \dots, 12\!-
14$.

Frequencies are 1,2,1,5,6,2,31,2,1,5,6,2,3.

∑fi=20 \sum f_i = 20

∑fixi=1⋅1+2⋅3+1⋅5+5⋅7+6⋅9+2⋅11+3⋅13 \sum f_i x_i = 1\cdot1 + 2\cdot3 + 1\cdot5 + 5\cdot7 + 6\cdot9 + 2\cdot11 + 3\cdot13

=1+6+5+35+54+22+39=162 = 1+6+5+35+54+22+39 = 162

So,

xˉ=∑fixi∑fi=16220=8.1 \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1

But the chapter's printed exercise table is for class intervals and frequencies; the mean number of plants per house comes from the same data as shown there. Using the grouped-data calculation with class marks 1,3,5,7,9,11,131,3,5,7,9,11,13 gives the mean 8.1 plants per house.

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2Which method did you use for finding the mean, and why?Show solution
I would use the direct method, because the class intervals and frequencies are small and the calculation is simple. The mean can be found directly from xˉ=∑fixi∑fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i} without needing any simplification.

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2Consider the following distribution of daily wages of 50 workers of a factory.Show solution
Use the direct method or the step-deviation method. Since the class intervals are equal and the numbers are manageable, the direct method is appropriate; if we want shorter calculations, the step-deviation method is also suitable.

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4Find the mean daily wages of the workers of the factory by using an appropriate method.Show solution
Take class marks xix_i as 510,530,550,570,590510, 530, 550, 570, 590.

Frequencies fif_i are 12,14,8,6,1012, 14, 8, 6, 10.

∑fi=50 \sum f_i = 50

∑fixi=12⋅510+14⋅530+8⋅550+6⋅570+10⋅590 \sum f_i x_i = 12\cdot510 + 14\cdot530 + 8\cdot550 + 6\cdot570 + 10\cdot590

=6120+7420+4400+3420+5900=27260 = 6120 + 7420 + 4400 + 3420 + 5900 = 27260

xˉ=2726050=545.2 \bar{x} = \frac{27260}{50} = 545.2

But the printed textbook answer for this exercise is based on the grouped distribution given in the chapter examples and uses the appropriate method to obtain the mean daily wages as ₹545.2. If the options do not include this value, the computed answer must be taken.

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3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency ff.Show solution
For the class intervals $11\!-
13, 13\!-
15, 15\!-
17, 17\!-
19, 19\!-
21, 21\!-
23, 23\!-
25$, the class marks are

12,14,16,18,20,22,2412, 14, 16, 18, 20, 22, 24.

Frequencies are 7,6,9,13,f,5,47, 6, 9, 13, f, 5, 4 and mean is given as 1818.

Using
xˉ=∑fixi∑fi \bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Compute known products:
7⋅12=84, 6⋅14=84, 9⋅16=144, 13⋅18=234, 5⋅22=110, 4⋅24=96 7\cdot12=84,\ 6\cdot14=84,\ 9\cdot16=144,\ 13\cdot18=234,\ 5\cdot22=110,\ 4\cdot24=96

So,
∑fixi=752+20f \sum f_i x_i = 752 + 20f

and
∑fi=44+f \sum f_i = 44 + f

Given mean =18=18:
18=752+20f44+f 18 = \frac{752 + 20f}{44 + f}

752+20f=792+18f 752 + 20f = 792 + 18f

2f=40 2f = 40

f=20 f = 20

The computed missing frequency is 20. If the printed options differ, the computed answer should be used.

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4Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.Show solution
Use the direct method with class marks:

66.5,69.5,72.5,75.5,78.5,81.5,84.566.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5

Frequencies:

2,4,3,8,7,4,22,4,3,8,7,4,2

∑fi=30 \sum f_i = 30

∑fixi=2(66.5)+4(69.5)+3(72.5)+8(75.5)+7(78.5)+4(81.5)+2(84.5) \sum f_i x_i = 2(66.5)+4(69.5)+3(72.5)+8(75.5)+7(78.5)+4(81.5)+2(84.5)

=133+278+217.5+604+549.5+326+169=2277 =133+278+217.5+604+549.5+326+169=2277

xˉ=227730=75.9 \bar{x} = \frac{2277}{30} = 75.9

So the mean heartbeats per minute is 75.9.

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5In 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.Show solution
Find the mean using the direct method with class marks of the boxes and then decide the appropriate method. The data has large frequencies, so the step-deviation method is convenient.

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8Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?Show solution
Class marks are 51,54,57,60,6351, 54, 57, 60, 63.

Frequencies are 15,110,135,115,2515, 110, 135, 115, 25.

∑fi=400 \sum f_i = 400

∑fixi=15⋅51+110⋅54+135⋅57+115⋅60+25⋅63 \sum f_i x_i = 15\cdot51 + 110\cdot54 + 135\cdot57 + 115\cdot60 + 25\cdot63

=765+5940+7695+6900+1575=22875 = 765 + 5940 + 7695 + 6900 + 1575 = 22875

xˉ=22875400=57.1875 \bar{x} = \frac{22875}{400} = 57.1875

So the mean number of mangoes per packing box is 57.1875, i.e. about 57.19.

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6The table below shows the daily expenditure on food of 25 households in a locality.Show solution
Since the expenditure classes and frequencies are small, the direct method is suitable. Take class marks of each interval and compute the mean from xˉ=∑fixi∑fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}.

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10Find the mean daily expenditure on food by a suitable method.Show solution
Class marks are 125,175,225,275,325125, 175, 225, 275, 325.

Frequencies are 4,5,12,2,24,5,12,2,2.

∑fi=25 \sum f_i = 25

∑fixi=4⋅125+5⋅175+12⋅225+2⋅275+2⋅325 \sum f_i x_i = 4\cdot125 + 5\cdot175 + 12\cdot225 + 2\cdot275 + 2\cdot325

=500+875+2700+550+650=5275 = 500 + 875 + 2700 + 550 + 650 = 5275

xˉ=527525=211 \bar{x} = \frac{5275}{25} = 211

So the mean daily expenditure on food is ₹211.

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7To 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:Show solution
Use the step-deviation method since the values are small but it still simplifies calculation. Form class marks and compute the mean concentration from the frequencies.

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12Find the mean concentration of SO₂ in the air.Show solution
Class marks are 0.02,0.06,0.10,0.14,0.18,0.220.02, 0.06, 0.10, 0.14, 0.18, 0.22.

Frequencies are 4,9,9,2,4,24,9,9,2,4,2.

∑fi=30 \sum f_i = 30

∑fixi=4(0.02)+9(0.06)+9(0.10)+2(0.14)+4(0.18)+2(0.22) \sum f_i x_i = 4(0.02)+9(0.06)+9(0.10)+2(0.14)+4(0.18)+2(0.22)

=0.08+0.54+0.90+0.28+0.72+0.44=2.96 =0.08+0.54+0.90+0.28+0.72+0.44=2.96

xˉ=2.9630=0.0987≈0.10 \bar{x}=\frac{2.96}{30}=0.0987\approx 0.10

So the mean concentration of SO2SO_2 is 0.10 ppm.

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8A 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.Show solution
Class marks are 3,8,12,17,24,33,393,8,12,17,24,33,39 for the intervals $0\!-
6, 6\!-
10, 10\!-
14, 14\!-
20, 20\!-
28, 28\!-
38, 38\!-
40$.

Frequencies are 11,10,7,4,4,3,111,10,7,4,4,3,1.

∑fi=40 \sum f_i = 40

∑fixi=11⋅3+10⋅8+7⋅12+4⋅17+4⋅24+3⋅33+1⋅39 \sum f_i x_i = 11\cdot3 + 10\cdot8 + 7\cdot12 + 4\cdot17 + 4\cdot24 + 3\cdot33 + 1\cdot39

=33+80+84+68+96+99+39=499 =33+80+84+68+96+99+39=499

xˉ=49940=12.475 \bar{x} = \frac{499}{40} = 12.475

So the mean number of days absent is 12.475, about 12.5 days.

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9The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.Show solution
Class marks are 50,60,70,80,9050,60,70,80,90.

Frequencies are 3,10,11,8,33,10,11,8,3.

∑fi=35 \sum f_i = 35

∑fixi=3⋅50+10⋅60+11⋅70+8⋅80+3⋅90 \sum f_i x_i = 3\cdot50 + 10\cdot60 + 11\cdot70 + 8\cdot80 + 3\cdot90

=150+600+770+640+270=2430 =150+600+770+640+270=2430

xˉ=243035=69.43 \bar{x}=\frac{2430}{35}=69.43

So the mean literacy rate is 69.43%.

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EXERCISE 13.2

1The following table shows the ages of the patients admitted in a hospital during a year:Show solution
First, convert the data into a grouped frequency table with continuous class intervals, then use the median formula and mode formula for grouped data.

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2Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.Show solution
The mode is the class with the highest frequency, and the mean is found by taking all class marks into account. For these ages, the modal class is the class with the greatest number of patients. The mean gives the average age of patients, while the mode gives the most common age group. A comparison of the two tells us about the typical age and the most frequent age of admission.

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2The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :Show solution
Find the modal class from the class with the highest frequency, then apply the grouped-data mode formula:

Mode=l+(f1−f02f1−f0−f2)h \text{Mode} = l + \left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h

This gives the most common lifetime interval.

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4Determine the modal lifetimes of the components.
3The 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 :
4The 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.
5The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
8Find the mode of the data.
6A 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 :

EXERCISE 13.3

1The 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.
2If the median of the distribution given below is 28.5, find the values of xx and yy.
3A 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.
4The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
5Find the median length of the leaves.
5The following table gives the distribution of the life time of 400 neon lamps :
7Find the median life time of a lamp.
6100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
9Determine 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.
7The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

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