NCERT Class 10 Math Chapter 13 Important Questions Answer – Statistics

Class10
Chapter Statistics
SubjectMath
CategoryImportant Question Answer

Class 10 Math Chapter 13 Important Question Answer

Q1. 100 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:

Number of letters1-44-77-1010-1313-1616-19
Number of surnames 630401644

Find the median numbers of letters in the surnames.

Ans.

Number of lettersNumber of surnamesCumulative Frequency
1-4

 

4-7

 

7-10

 

10-13

 

13-16

 

16-19

6

 

30

 

40

 

16

 

4

 

4

6

 

36

 

76

 

92

 

96

 

100

 

Now, n=100, \displaystyle \frac{n}{2}=\frac{{100}}{2}=50. This observation lies in the class 7-10.

Then, l (lower limit) = 7,

cf (the cumulative frequency of the class preceding 7-10) = 36

f (the frequency of the median class 7-10) = 40,

h (the class size) = 3.

Using the formula, Median = l + \displaystyle \left( {\frac{{\frac{n}{2}-cf}}{f}} \right) × h

Median =7 + \displaystyle \left( {\frac{{50-36}}{{40}}} \right)\times 3 =7  + 1.05 = 9.05

Q2. The median of the following data is 28.5. Find the values of x and y, if the total frequency is 60.

Class interval0-1010-2020-3030-4040-5050-60
Frequency5x2015y5

 

Ans.

Class-IntervalFrequencyCumulative Frequency
0-10

 

10-20

 

20-30

 

30-40

 

40-50

 

50-60

 

5

 

x

 

20

 

15

 

y

 

5

5

 

5 + x

 

25 + x

 

40 + x

 

40 + x + y

 

45 + x + y

It is given that n = 60

So, 45 + x + y = 60  i.e., x + y = 15

The Median is 28.5, which lies in the class 20-30

So, l = 20,  f = 20,  cf = 5 + x,  h = 10

Using the formula, Median = l + \displaystyle \left( {\frac{{\frac{n}{2}-cf}}{f}} \right) × h

28.5 = 20 + \displaystyle \left( {\frac{{30-5-x}}{{20}}} \right)\times 10

28.5  – 20 = \displaystyle \frac{{25-x}}{2}

17 = 25 – x

x = 8

putting vlaue of x in x + y = 15,

we get y = 7

Q3. Consider the distribution of daily wages of 50 workers of a factory :

Daily Wages (in Rs.)100-120 120-140140-160160-180180-200
Number of workers12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans.

As the given data is large, we will use step deviation method to find the mean.

Here a = 150, h = 20

Class IntervalFrequency

(fi)

Class Marks

(ui)

 \displaystyle {{u}_{i}}=\frac{{{{x}_{i}}-a}}{h}fiui
100-120

120-140

140-160

160-180

180-200

12

14

8

6

10

110

130

150 (a)

170

190

-2

-1

0

1

2

 

-24

-14

0

6

20

\displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}}}=50 \displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}}}{{u}_{i}}=-12

 

Step Deviation Method : \displaystyle \overline{x} = a + × h

= 150 + \displaystyle 20\left( {\frac{{-12}}{{50}}} \right) = 150 – \displaystyle \frac{{240}}{{50}}

= 145.20

Hence, the mean daily expenditure on food is Rs.145.20

Q4. 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.

Daily pocket (in Rs.)11-1313-1515-1717-1919-2121-2323-25
Number of chapter76913f54

Ans.

Daily pocket allowances ( in Rs.)Class Mark

(xi)

Number of children

(fi)

Di = xi – 18fidi
11-13

13-15

15-17

17-19

19-21

21-23

23-25

12

14

16

18 ( let a = 18)

20

22

24

7

6

9

13

f

5

4

-6

-4

-2

0

2

4

6

-42

-24

-18

0

2f

20

24

Total \displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}=44+f}} \displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}{{d}_{i}}=2f-40}}

We have,

Assumed Mean Method : \displaystyle \overline{x} = a +

18 = 18 + \displaystyle \frac{{2f-40}}{{44+f}}

0 = \displaystyle \frac{{2f-40}}{{44+f}}

2f – 40 = 0

f = 20

Therefore, required frequency is 20.

Q5. The following data gives the information on the life-time (in hours) of 75 electrical instruments:

Life-time (in hours)0-2020-4040-6060-8080-100100-120
Frequency1015122189

Determine the modal lifetime of the components.

Ans.

Here the maximum frequency is 21. So the modal class is 60-80.

Now,

Modal class = 60-80, lower limit (l) of modal class = 60, class size (h) = 20

Frequency (f1) of modal class = 21

Frequency (f0) of class preceding the modal class = 12,

Frequency (f2) of class succeeding the modal class = 8

Now, Using Mode = l + × h

Mode = 60 + \displaystyle \left( {\frac{{21-12}}{{2(21)-12-8}}} \right)\times 20 = 60 + \displaystyle \frac{{90}}{{11}}= 68.18

Q6. The median of the following data is 525. Find the value of x and y, if the total frequency is 100:

Class-IntervalFrequency
0-100

100-200

200-300

300-400

400-500

500-600

600-700

700-800

800-900

900-1000

2

5

x

12

17

20

y

9

7

4

Ans.

Class-IntervalFrequencyCumulative Frequency
0-100

 

100-200

 

200-300

 

300-400

 

400-500

 

500-600

 

600-700

 

700-800

 

800-900

 

900-1000

2

 

5

 

x

 

12

 

17

 

20

 

y

 

9

 

7

 

4

2

 

7

 

7+x

 

19 + x

 

36 + x

 

56 + x

 

56 + x + y

 

65 + x + y

 

72 + x + y

 

76 + x + y

 

 It is given that n = 100

So, 76 + x + y = 100  i.e., x + y = 24

The Median is 525, which lies in the class 500-600

So, l = 500,  f = 20,  cf = 36 + x,  h = 100

Using the formula, Median = l + \displaystyle \left( {\frac{{\frac{n}{2}-cf}}{f}} \right) × h

525 = 500 + \displaystyle \left( {\frac{{50-36-x}}{{20}}} \right)\times 100

525 – 500 = (14 – x)×5

25 = 70 – 5x

5x = 70 – 25 = 45

x = 9

putting value of x in x + y = 24,

we get 9 + y = 24

y = 15

Q7. A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and the following data was :

Obtained height (in cm)Less than 140Less than 145Less than 150Less than 155Less than 160Less than 165
Number of girls41129404651

Find the median height.

Ans.

The given distribution being of the less than type. Therefore, the classes should be below 140, 140-145, 145-150, ….., 160-165.

The frequency of class interval below 140 is 4, frequency of class interval 140-145 is 11-4=7. Similarly the frequency of 145-150 is 29-11=18 and so on.

So, our frequency distribution table with the given cumulative frequencies becomes :

Class IntervalFrequencyCumulative Frequency  (cf)
Below 14044
140-145711
145-1501829
150-1551140
155-160646
160-165551

Now, n=51, \displaystyle \frac{n}{2}=\frac{{51}}{2}=25.5. This observation lies in the class 145-150.

Then, l (lower limit) = 145,

cf (the cumulative frequency of the class preceding 145-150) = 11

f (the frequency of the median class 145-150) = 18,

h (the class size) = 5.

Using the formula, Median = l + \displaystyle \left( {\frac{{\frac{n}{2}-cf}}{f}} \right) × h

Median = 145 + \displaystyle \left( {\frac{{25.5-11}}{{18}}} \right)\times 5

= 145 + \displaystyle \frac{{72.5}}{{18}} = 149.03

Q8. The table below shows daily expenditure on food of 25 households in a locality :

Daily Expenditure (in Rs.)100-150150-200200-250250-300300-350
No. of households.451222

Find the mean daily expenditure on food by suitable method.

Ans.

Here, a = 225 and h = 50

Class IntervalFrequencyClass Marks \displaystyle {{u}_{i}}=\frac{{{{x}_{i}}-a}}{h}fiui
100-150

150-200

200-250

250-300

300-350

4

5

12

2

2

125

175

225 (a)

275

325

-2

-1

0

1

2

 

-8

-5

0

2

4

\displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}}}=25 \displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}}}{{u}_{i}}=-7

Step Deviation Method : \displaystyle \overline{x} = a + × h

= 225 + \displaystyle 50\left( {\frac{{-7}}{{25}}} \right) = 225 – 14

= 211.

Hence, the mean daily expenditure on food is Rs.211.

Q9. The following distribution shows the daily pocket money of children of a school:

Daily Pocket Money (Rs.)11-13 13-15 15-1717-19 19-2121-2323-25
Number of Children769132054

 Find the average daily pocket money of children.

Ans.

Daily pocket allowances ( in Rs.)Class Mark

(xi)

Number of children

(fi)

Di = xi – 18fidi
11-13

13-15

15-17

17-19

19-21

21-23

23-25

12

14

16

18 ( let a = 18)

20

22

24

7

6

9

13

20

5

4

-6

-4

-2

0

2

4

6

-42

-24

-18

0

40

20

24

Total \displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}=64}} \displaystyle \sum\limits_{{}}^{{}}{{{{f}_{i}}{{d}_{i}}=0}}

 

We have,

Assumed Mean Method : \displaystyle \overline{x} = a +

\displaystyle \overline{x} = 18 + \displaystyle \frac{0}{{64}} = 18

Therefore, the average daily pocket money of children is Rs. 18.

Q10. The following distribution gives the monthly consumption of consumers of a locality. Find the median of the distribution.

Monthly consumption (in units)65-8585-105105-125125-145145-165165-185
Number of consumers481320144

Ans.

Monthly consumption

 (in units)

Number of consumers

(f)

Cumulative Frequency

 (cf)

65-8544
85-105812
105-1251325
125-1452045
145-1651459
165-185463

Now, n = 63, \displaystyle \frac{n}{2}=\frac{{63}}{2}=31.5. This observation lies in the class 125-145 .

Then, l (lower limit) = 125,

cf (the cumulative frequency of the class preceding 125-145) = 25,

f (the frequency of the median class 125-145) = 20,

h (the class size) = 20.

Using the Formula, Median = l + \displaystyle \left( {\frac{{\frac{n}{2}-cf}}{f}} \right) × h

Median = 125 + \displaystyle \left( {\frac{{31.5-25}}{{20}}} \right)20 = 125 + 6.5 = 126.5

Q11. The wickets taken by a bowler in 10 cricket matches are as follows:
3    5     2    1     2     0    5     1     2     4
Find the mode of the data.

Ans. Arranging data in ascending order : 0, 1, 2, 2, 2, 3, 4, 5, 5

From above data we can clearly conclude that 3 is the mode of data as it is used in max times.

Therefore,

Mode = 3

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