NCERT Solutions: Statistics

# NCERT Solutions Class 11 Maths Chapter 13 - Statistics

## Exercise - 13.1

Q1: Find the mean deviation about the mean for the data   4, 7, 8, 9, 10, 12, 13, 17
Ans: The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data,
The deviations of the respective observations from the mean   are
–6, – 3, –2, –1, 0, 2, 3, 7
The absolute values of the deviations, i.e.  , are
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is

Q2: Find the mean deviation about the mean for the data  38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Ans: The given data is
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data,

The deviations of the respective observations from the mean   are
–12, 20, –2, –10, –8, 5, 13, –4, 4, –6
The absolute values of the deviations, i.e.   , are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
The required mean deviation about the mean is

Q3: Find the mean deviation about the median for the data.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Ans: The given data is
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Here, the numbers of observations are 12, which is even.
Arranging the data in ascending order, we obtain
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e.  are
–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The absolute values of the deviations,  , are
3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The required mean deviation about the median is

Q4: Find the mean deviation about the median for the data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Ans: The given data is
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Here, the number of observations is 10, which is even.
Arranging the data in ascending order, we obtain
36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e.  are
–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The absolute values of the deviations,  , are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Thus, the required mean deviation about the median is

Q5: Find the mean deviation about the mean for the data.

Ans:

Q6: Find the mean deviation about the mean for the data

Ans:

Q7: Find the mean deviation about the median for the data.

Ans: The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

Here, N = 26, which is even.
Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e.  are

and

Q8: Find the mean deviation about the median for the data

Ans: The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

Here, N = 29, which is odd.

observation = 15th observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
∴ Median = 30
The absolute values of the deviations from median, i.e.  are

∴

Q9: Find the mean deviation about the mean for the data.

Ans: The following table is formed

Here,

Q10: Find the mean deviation about the mean for the data

Ans: The following table is formed.

Here,

Q11: Find the mean deviation about median for the following data:

Ans: The following table is formed.

The class interval containing the  or 25th item is 20 – 30.
Therefore, 20 – 30 is the median class.
It is known that,

Here, l = 20, C = 14, f = 14, h = 10, and N = 50
∴ Median =
Thus, mean deviation about the median is given by,

Q12: Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Ans: The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
The table is formed as follows.

 Age Number fi Cumulative frequency (c.f.) Mid-point xi |xi – Med.| fi |xi – Med.| 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 100 735

The class interval containing the  or 50th item is 35.5 – 40.5.
Therefore, 35.5 – 40.5 is the median class.
It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,

## Exercise - 13.2

Q1: Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Ans: 6, 7, 10, 12, 13, 4, 8, 12
Mean,
The following table is obtained.

Q2: Find the mean and variance for the first n natural numbers
Ans: The mean of first n natural numbers is calculated as follows.

Q3: Find the mean and variance for the first 10 multiples of 3
Ans: The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10

The following table is obtained.

Q4: Find the mean and variance for the data

Ans: The data is obtained in tabular form as follows.

 xi f i fixi 6 2 12 –13 169 338 10 4 40 –9 81 324 14 7 98 –5 25 175 18 12 216 –1 1 12 24 8 192 5 25 200 28 4 112 9 81 324 30 3 90 11 121 363 40 760 1736

Here, N = 40,

Q5: Find the mean and variance for the data

Ans: The data is obtained in tabular form as follows.

 xi f i fixi 92 3 276 –8 64 192 93 2 186 –7 49 98 97 3 291 –3 9 27 98 2 196 –2 4 8 102 6 612 2 4 24 104 3 312 4 16 48 109 3 327 9 81 243 22 2200 640

Here, N = 22,

Q6: Find the mean and standard deviation using short-cut method.

Ans: The data is obtained in tabular form as follows.

 xi fi yi2 fiyi fiyi2 60 2 –4 16 –8 32 61 1 –3 9 –3 9 62 12 –2 4 –24 48 63 29 –1 1 –29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 100 220 0 286

Mean,

Q7: Find the mean and variance for the following frequency distribution.

 Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2

Ans:

 Class Frequency fi Mid-point xi yi2 fiyi fiyi2 0-30 2 15 –3 9 –6 18 30-60 3 45 –2 4 –6 12 60-90 5 75 –1 1 –5 5 90-120 10 105 0 0 0 0 120-150 3 135 1 1 3 3 150-180 5 165 2 4 10 20 180-210 2 195 3 9 6 18 30 2 76

Mean,

Q8: Find the mean and variance for the following frequency distribution.

 Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6

Ans:

 Class Frequency fi Mid-point xi yi2 fiyi fiyi2 0-10 5 5 –2 4 –10 20 10-20 8 15 –1 1 –8 8 20-30 15 25 0 0 0 0 30-40 16 35 1 1 16 16 40-50 6 45 2 4 12 24 50 10 68

Mean,

Q9: Find the mean, variance and standard deviation using short-cut method

Ans:

 Class Interval Frequency fi Mid-point xi yi2 fiyi fiyi2 70-75 3 72.5 –4 16 –12 48 75-80 4 77.5 –3 9 –12 36 80-85 7 82.5 –2 4 –14 28 85-90 7 87.5 –1 1 –7 7 90-95 15 92.5 0 0 0 0 95-100 9 97.5 1 1 9 9 100-105 6 102.5 2 4 12 24 105-110 6 107.5 3 9 18 54 110-115 3 112.5 4 16 12 48 60 6 254

Mean,

Q10: The diameters of circles (in mm) drawn in a design are given below:

Ans:

 Class Interval Frequency fi Mid-point xi fi2 fiyi fiyi2 32.5-36.5 15 34.5 –2 4 –30 60 36.5-40.5 17 38.5 –1 1 –17 17 40.5-44.5 21 42.5 0 0 0 0 44.5-48.5 22 46.5 1 1 22 22 48.5-52.5 25 50.5 2 4 50 100 100 25 199

Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.

Mean,

Old NCERT Questions

Q1: From the data given below state which group is more variable, A or B?

 Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7

Ans: Firstly, the standard deviation of group A is calculated as follows.

 Marks Group A fi Mid-point xi yi2 fiyi fiyi2 10-20 9 15 –3 9 –27 81 20-30 17 25 –2 4 –34 68 30-40 32 35 –1 1 –32 32 40-50 33 45 0 0 0 0 50-60 40 55 1 1 40 40 60-70 10 65 2 4 20 40 70-80 9 75 3 9 27 81 150 –6 342

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

 Marks Group Bfi Mid-pointxi yi2 fiyi fiyi2 10-20 10 15 –3 9 –30 90 20-30 20 25 –2 4 –40 80 30-40 30 35 –1 1 –30 30 40-50 25 45 0 0 0 0 50-60 43 55 1 1 43 43 60-70 15 65 2 4 30 60 70-80 7 75 3 9 21 63 150 –6 366

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.

Q2: From the prices of shares X and Y below, find out which is more stable in value:

 X 35 54 52 53 56 58 52 50 51 49 Y 108 107 105 105 106 107 104 103 104 101

Ans: The prices of the shares X are
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10

The following table is obtained corresponding to shares X.

 xi 35 –16 256 54 3 9 52 1 1 53 2 4 56 5 25 58 7 49 52 1 1 50 –1 1 51 0 0 49 –2 4 350

The prices of share Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

 yi 108 3 9 107 2 4 105 0 0 105 0 0 106 1 1 107 2 4 104 –1 1 103 –2 4 104 –1 1 101 –4 16 40

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.

Q3: An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 Firm A Firm B No. of wage earners 586 648 Mean of monthly wages Rs 5253 Rs 5253 Variance of the distribution of wages 100 121
• Which firm A or B pays larger amount as monthly wages?
• Which firm, A or B, shows greater variability in individual wages?

Ans: (i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
∴Total amount paid = Rs 5253 × 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 × 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.
(ii) Variance of the distribution of wages in firm A   = 100
∴ Standard deviation of the distribution of wages in firm
A ((σ1) =
Variance of the distribution of wages in firm   = 121
∴ Standard deviation of the distribution of wages in firm
The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.
Thus, firm B has greater variability in the individual wages.

Q4: The following is the record of goals scored by team A in a football session:

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Ans: The mean and the standard deviation of goals scored by team A are calculated as follows.

 No. of goals scored No. of matches fixi xi2 fixi2 0 1 0 0 0 1 9 9 1 9 2 7 14 4 28 3 5 15 9 45 4 3 12 16 48 25 50 130

Thus, the mean of both the teams is same.

The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.
Thus, team A is more consistent than team B.

Q5: The sum and sum of squares corresponding to length x (in cm) and weight (in gm) of 50 plant products are given below:

Which is more varying, the length or weight?
Ans:

Here, N = 50

∴ Mean,

Mean,

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

Q6: The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Ans: Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

From (1), we obtain
x2 +  y2 + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x2 +  y2 – 2xy = 80 – 64 = 16
xy = ± 4 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when xy = 4
x = 4 and y = 8, when xy = –4
Thus, the remaining observations are 4 and 8.

Q7: The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Ans:
Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

From (1), we obtain
x2 +  y2 + 2xy = 196 … (3)
From (2) and (3), we obtain
2xy = 196 – 100
⇒ 2xy = 96 … (4)
Subtracting (4) from (2), we obtain
x2 +  y2 – 2xy = 100 – 96
⇒ (xy)2 = 4
xy = ± 2 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when xy = 2
x = 6 and y = 8 when xy = – 2
Thus, the remaining observations are 6 and 8.

Q8: The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Ans: Let the observations be x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are yi, then

From (1) and (2), it can be observed that,

Substituting the values of xi and    in (2), we obtain

Therefore, variance of new observations =
Hence, the standard deviation of new observations is

Q9: Given that   is the mean and σ2 is the variance of n observations x1, x2xn. Prove that the mean and variance of the observations ax1, ax2, ax3axn are    and a2 σ2, respectively (a ≠ 0).
Ans:
The given n observations are x1, x2xn.

Mean =

Variance = σ2

If each observation is multiplied by a and the new observations are yi, then

Therefore, mean of the observations, ax1, ax2axn, is   .

Substituting the values of xiand   in (1), we obtain

Thus, the variance of the observations, ax1, ax2axn, is a2 σ2.

Q10: The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
Ans: (i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2

That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean

(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
∴ Correct sum of observations = 200 – 8 + 12 = 204

Q11: The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Which of the three subjects shows the highest variability in marks and which shows the lowest?
Ans: Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coefficient of variation (C.V.) is given by   .

The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Q12: The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Ans: Number of observations (n) = 100
Incorrect mean
Incorrect standard deviation

∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

Standered deviation

Correct Standered deviation

The document NCERT Solutions Class 11 Maths Chapter 13 - Statistics is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

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## FAQs on NCERT Solutions Class 11 Maths Chapter 13 - Statistics

 1. What are the different types of statistics?
Ans. Descriptive statistics and inferential statistics are the two main types of statistics. Descriptive statistics involve organizing, summarizing, and presenting data, while inferential statistics involve making predictions or inferences about a population based on a sample.
 2. How is central tendency calculated in statistics?
Ans. Central tendency in statistics is typically measured using the mean, median, and mode. The mean is calculated by adding up all the values and dividing by the number of values. The median is the middle value when the data is arranged in order, and the mode is the most frequently occurring value.
 3. What is the importance of statistics in research?
Ans. Statistics plays a crucial role in research by helping researchers analyze and interpret data. It allows researchers to draw meaningful conclusions, make predictions, and test hypotheses based on the data collected.
 4. How can outliers affect statistical analysis?
Ans. Outliers are data points that significantly differ from the rest of the data. They can skew the results of statistical analysis, leading to inaccurate conclusions. It is important to identify and address outliers to ensure the reliability of the analysis.
 5. What are the different measures of variability in statistics?
Ans. The range, variance, and standard deviation are common measures of variability in statistics. The range is the difference between the highest and lowest values, while variance and standard deviation measure the spread or dispersion of data around the mean.

## Mathematics (Maths) for JEE Main & Advanced

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