#### Chapters

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Values of Trigonometric function at sum or difference of angles

Chapter 8: Transformation formulae

Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle

Chapter 10: Sine and cosine formulae and their applications

Chapter 11: Trigonometric equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 14: Quadratic Equations

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progression

Chapter 20: Geometric Progression

Chapter 21: Some special series

Chapter 22: Brief review of cartesian system of rectangular co-ordinates

Chapter 23: The straight lines

Chapter 24: The circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to three dimensional coordinate geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical reasoning

Chapter 32: Statistics

Chapter 33: Probability

#### RD Sharma Mathematics Class 11

## Chapter 32 : Statistics

#### Page 6

Calculate the mean deviation about the median of the observation:

3011, 2780, 3020, 2354, 3541, 4150, 5000

Calculate the mean deviation about the median of the observation:

38, 70, 48, 34, 42, 55, 63, 46, 54, 44

Calculate the mean deviation about the median of the observation:

34, 66, 30, 38, 44, 50, 40, 60, 42, 51

Calculate the mean deviation about the median of the observation:

22, 24, 30, 27, 29, 31, 25, 28, 41, 42

Calculate the mean deviation about the median of the observation:

38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Calculate the mean deviation from the mean for the data:

4, 7, 8, 9, 10, 12, 13, 17

Calculate the mean deviation from the mean for the data:

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Calculate the mean deviation from the mean for the data:

38, 70, 48, 40, 42, 55, 63, 46, 54, 44a

Calculate the mean deviation from the mean for the data:

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Calculate the mean deviation of the following income groups of five and seven members from their medians:

IIncome in Rs. |
IIIncome in Rs. |

4000 4200 4400 4600 4800 |
300 4000 4200 4400 4600 4800 5800 |

The lengths (in cm) of 10 rods in a shop are given below:

40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2

Find mean deviation from median

The lengths (in cm) of 10 rods in a shop are given below:

40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2

Find mean deviation from the mean also.

In 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 find the number of observations lying between

\[\bar{ X } \] + M.D, where M.D. is the mean deviation from the mean.

In 22, 24, 30, 27, 29, 31, 25, 28, 41, 42 find the number of observations lying between

\[\bar { X } \] − M.D. and

\[\bar { X } \] + M.D, where M.D. is the mean deviation from the mean.

In 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 find the number of observations lying between

\[\bar { X } \] − M.D. and

\[\bar { X } \] + M.D, where M.D. is the mean deviation from the mean.

#### Page 11

Calculate the mean deviation from the median of the following frequency distribution:

Heights in inches | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |

No. of students | 15 | 20 | 32 | 35 | 35 | 22 | 20 | 10 | 8 |

The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution:

Number of calls | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Frequency | 14 | 21 | 25 | 43 | 51 | 40 | 39 | 12 |

Compute the mean deviation about median.

Calculate the mean deviation about the median of the following frequency distribution:

x_{i} |
5 | 7 | 9 | 11 | 13 | 15 | 17 |

f_{i} |
2 | 4 | 6 | 8 | 10 | 12 | 8 |

Find the mean deviation from the mean for the data:

x_{i} |
5 | 7 | 9 | 10 | 12 | 15 |

f_{i} |
8 | 6 | 2 | 2 | 2 | 6 |

Find the mean deviation from the mean for the data:

x_{i} |
5 | 10 | 15 | 20 | 25 |

f_{i} |
7 | 4 | 6 | 3 | 5 |

Find the mean deviation from the mean for the data:

x_{i} |
10 | 30 | 50 | 70 | 90 |

f_{i} |
4 | 24 | 28 | 16 | 8 |

Find the mean deviation from the mean for the data:

Size |
20 | 21 | 22 | 23 | 24 |

Frequency |
6 | 4 | 5 | 1 | 4 |

Find the mean deviation from the mean for the data:

Size |
1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |

Frequency |
3 | 3 | 4 | 14 | 7 | 4 | 3 | 4 |

Find the mean deviation from the median for the data:

x_{i} |
15 | 21 | 27 | 30 | 35 |

f_{i} |
3 | 5 | 6 | 7 | 8 |

Find the mean deviation from the median for the data:

x_{i} |
74 | 89 | 42 | 54 | 91 | 94 | 35 |

f_{i} |
20 | 12 | 2 | 4 | 5 | 3 | 4 |

#### Pages 16 - 17

Compute the mean deviation from the median of the following distribution:

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequency | 5 | 10 | 20 | 5 | 10 |

Find the mean deviation from the mean for the data:

Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |

Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |

Find the mean deviation from the mean for the data:

Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |

Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |

Find the mean deviation from the mean for the data:

Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequencies | 6 | 8 | 14 | 16 | 4 | 2 |

Compute mean deviation from mean of the following distribution:

Mark | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |

No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |

The age distribution of 100 life-insurance policy holders is as follows:

Age (on nearest birth day) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |

No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |

Calculate the mean deviation from the median age

Find the mean deviation from the mean and from median of the following distribution:

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

No. of students | 5 | 8 | 15 | 16 | 6 |

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

Age: | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |

Number of persons | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |

Calculate the mean deviation about the mean for the following frequency distribution:

Class interval: | 0–4 | 4–8 | 8–12 | 12–16 | 16–20 |

Frequency | 4 | 6 | 8 | 5 | 2 |

Calculate mean deviation from the median of the following data:

Class interval: | 0–6 | 6–12 | 12–18 | 18–24 | 24–30 |

Frequency | 4 | 5 | 3 | 6 | 2 |

#### Page 28

Find the mean, variance and standard deviation for the data:

2, 4, 5, 6, 8, 17.

Find the mean, variance and standard deviation for the data:

6, 7, 10, 12, 13, 4, 8, 12.

Find the mean, variance and standard deviation for the data:

227, 235, 255, 269, 292, 299, 312, 321, 333, 348.

Find the mean, variance and standard deviation for the data 15, 22, 27, 11, 9, 21, 14, 9.

The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.

The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

The mean and standard deviation of 6 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.

The mean and variance of 8 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.

For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

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.

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 were omitted.

Show that the two formulae for the standard deviation of ungrouped data

\[\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}\] and

\[\sigma' = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}\] are equivalent, where \[X = \frac{1}{n}\sum_{} x_i\]

#### Pages 37 - 38

Find the standard deviation for the following distribution:

x : |
4.5 | 14.5 | 24.5 | 34.5 | 44.5 | 54.5 | 64.5 |

f : |
1 | 5 | 12 | 22 | 17 | 9 | 4 |

Table below shows the frequency *f* with which *'x' *alpha particles were radiated from a diskette

x : |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

f : |
51 | 203 | 383 | 525 | 532 | 408 | 273 | 139 | 43 | 27 | 10 | 4 | 2 |

Calculate the mean and variance.

Find the mean, mode, S.D. and coefficient of skewness for the following data:

Year render: | 10 | 20 | 30 | 40 | 50 | 60 |

No. of persons (cumulative): | 15 | 32 | 51 | 78 | 97 | 109 |

Find the standard deviation for the following data:

x : |
3 | 8 | 13 | 18 | 23 |

f : |
7 | 10 | 15 | 10 | 6 |

#### Pages 0 - 42

Calculate the mean and S.D. for the following data:

Expenditure in Rs: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequency: | 14 | 13 | 27 | 21 | 15 |

Calculate the standard deviation for the following data:

Class: | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |

Frequency: | 9 | 17 | 43 | 82 | 81 | 44 | 24 |

Calculate the A.M. and S.D. for the following distribution:

Class: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency: | 18 | 16 | 15 | 12 | 10 | 5 | 2 | 1 |

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.

Calculate the mean, median and standard deviation of the following distribution:

Class-interval: | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 | 61-65 | 66-70 |

Frequency: | 2 | 3 | 8 | 12 | 16 | 5 | 2 | 3 |

Find the mean and variance of frequency distribution given below:

x:_{i} |
1 ≤ x < 3 |
3 ≤ x < 5 |
5 ≤ x < 7 |
7 ≤ x < 10 |

f:_{i} |
6 | 4 | 5 | 1 |

The weight of coffee in 70 jars is shown in the following table:

Weight (in grams): | 200–201 | 201–202 | 202–203 | 203–204 | 204–205 | 205–206 |

Frequency: | 13 | 27 | 18 | 10 | 1 | 1 |

Determine the variance and standard deviation of the above distribution.

Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Calculate the mean, variance and standard deviation of the following frequency distribution.

Class: | 1–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |

Frequency: | 11 | 29 | 18 | 4 | 5 | 3 |

#### Pages 47 - 49

Two plants *A* and *B* of a factory show following results about the number of workers and the wages paid to them

Plant A | Plant B | |

No. of workers | 5000 | 6000 |

Average monthly wages | Rs 2500 | Rs 2500 |

Variance of distribution of wages | 81 | 100 |

In which plant *A* or *B* is there greater variability in individual wages?

The means and standard deviations of heights ans weights of 50 students of a class are as follows:

Weights | Heights | |

Mean | 63.2 kg | 63.2 inch |

Standard deviation | 5.6 kg | 11.5 inch |

Which shows more variability, heights or weights?

Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

Calculate coefficient of variation from the following data:

Income (in Rs): | 1000-1700 | 1700-2400 | 2400-3100 | 3100-3800 | 3800-4500 | 4500-5200 |

No. of families: | 12 | 18 | 20 | 25 | 35 | 10 |

An analysis of the weekly 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 |

Average weekly wages | Rs 52.5 | Rs. 47.5 |

Variance of the |
100 |
121 |

distribution of wages |

(i) Which firm *A* or *B* pays out larger amount as weekly wages?

(ii) Which firm *A* or *B* has greater variability in individual wages?

The following are some particulars of the distribution of weights of boys and girls in a class:

Number | Boys | Girls |

100 | 50 | |

Mean weight | 60 kg | 45 kg |

Variance | 9 | 4 |

Which of the distributions is more variable?

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:

Subject | Mathematics | Physics | Chemistry |

Mean | 42 | 32 | 40.9 |

Standard Deviation | 12 | 15 | 20 |

Which of the three subjects shows the highest variability in marks and which shows the lowest?

From the data given below state which group is more variable, *G*_{1} or *G*_{2}?

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Group G_{1} |
9 | 17 | 32 | 33 | 40 | 10 | 9 |

Group G_{2} |
10 | 20 | 30 | 25 | 43 | 15 | 7 |

Find the coefficient of variation for the following data:

Size (in cms): | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |

No. of items: | 2 | 8 | 20 | 35 | 20 | 15 |

From the prices of shares *X* and *Y* given 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 |

Life of bulbs produced by two factories A and B are given below:

Length of life (in hours): |
550–650 | 650–750 | 750–850 | 850–950 | 950–1050 |

Factory A:(Number of bulbs) |
10 | 22 | 52 | 20 | 16 |

Factory B:(Number of bulbs) |
8 | 60 | 24 | 16 | 12 |

The bulbs of which factory are more consistent from the point of view of length of life?

Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests:

Ravi: | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 35 | 60 |

Hashina: | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 |

Who is more intelligent and who is more consistent?

#### Page 49

Write the variance of first *n* natural numbers.

If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.

If *x*_{1}, *x*_{2}, ..., *x _{n}* are

*n*values of a variable

*X*and

*y*

_{1},

*y*

_{2}, ...,

*y*are

_{n}*n*values of variable

*Y*such that

*y*=

_{i}*ax*+

_{i}*b*;

*i*= 1, 2, ...,

*n*, then write Var(

*Y*) in terms of Var(

*X*).

If *X* and *Y* are two variates connected by the relation

*X*) = σ

^{2}, then write the expression for the standard deviation of

*Y*.

In a series of 20 observations, 10 observations are each equal to *k* and each of the remaining half is equal to − *k*. If the standard deviation of the observations is 2, then write the value of *k*.

If each observation of a raw data whose standard deviation is σ is multiplied by *a*, then write the S.D. of the new set of observations.

If a variable *X* takes values 0, 1, 2,..., *n* with frequencies ^{n}C_{0}, ^{n}C_{1}, ^{n}C_{2} , ... , * ^{n}C_{n}*, then write variance

*X*.

#### Pages 50 - 52

For a frequency distribution mean deviation from mean is computed by

M.D. = \[\frac{\Sigma f}{\Sigma f \left| d \right|}\]

M.D. = \[\frac{\Sigma d}{\Sigma f}\]

M.D. = \[\frac{\Sigma f d}{\Sigma f}\]

M.D. = \[\frac{\Sigma f \left| d \right|}{\Sigma f}\]

For a frequency distribution standard deviation is computed by applying the formula

\[\sigma = \sqrt{\frac{\Sigma f d^2}{\Sigma f} - \left( \frac{\Sigma f d}{\Sigma f} \right)^2}\]

\[\sigma = \sqrt{\left( \frac{\Sigma f d}{\Sigma f} \right)^2 - \frac{\Sigma f d^2}{\Sigma f}}\]

\[\sigma = \sqrt{\frac{\Sigma f d^2}{\Sigma f} - \frac{\Sigma fd}{\Sigma f}}\]

\[\sqrt{\left( \frac{\Sigma fd}{\Sigma f} \right)^2 - \frac{\Sigma f d^2}{\Sigma f}}\]

If *v* is the variance and σ is the standard deviation, then

\[v = \frac{1}{\sigma^2}\]

\[v = \frac{1}{\sigma}\]

*v* = σ^{2}

*v*^{2} = σ

The mean deviation from the median is

equal to that measured from another value

maximum if all observations are positive

greater than that measured from any other value.

less than that measured from any other value.

If *n* = 10, \[X = 12\] and \[\Sigma x_i^2 = 1530\] , then the coefficient of variation is

36%

41%

25%

none of these

The standard deviation of the data:

x: |
1 | a |
a^{2} |
.... | a^{n} |

f: |
^{n}C_{0} |
^{n}C_{1} |
^{n}C_{2} |
.... | ^{n}C_{n} |

is

\[\left( \frac{1 + a^2}{2} \right)^n - \left( \frac{1 + a}{2} \right)^n\]

\[\left( \frac{1 + a^2}{2} \right)^{2n} - \left( \frac{1 + a}{2} \right)^n\]

\[\left( \frac{1 + a}{2} \right)^{2n} - \left( \frac{1 + a^2}{2} \right)^n\]

none of these

The mean deviation of the series *a*, *a* + *d*, *a* + 2*d*, ..., *a* + 2*n* from its mean is

\[\frac{(n + 1) d}{2n + 1}\]

\[\frac{nd}{2n + 1}\]

\[\frac{n (n + 1) d}{2n + 1}\]

\[\frac{(2n + 1) d}{n (n + 1)}\]

A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is

8.6

6.4

10.6

7.6

None of these

The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is

10 %

40 %

50 %

none of these

Let *x*_{1}, *x*_{2}, ..., *x _{n}* be values taken by a variable

*X*and

*y*

_{1},

*y*

_{2}, ...,

*y*be the values taken by a variable

_{n}*Y*such that

*y*=

_{i}*ax*+

_{i}*b*,

*i*= 1, 2,...,

*n*. Then,

Var (*Y*) = *a*^{2} Var (*X*)

Var (*X*) = *a*^{2} Var (*Y*)

Var (*X*) = Var (*X*) + *b*

none of these

If the standard deviation of a variable *X* is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] is

*a *σ

\[\frac{a}{c}\sigma\]

\[\left| \frac{a}{c} \right| \sigma\]

\[\frac{a\sigma + b}{c}\]

If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be

−4

−8

8

4

If two variates *X* and *Y* are connected by the relation \[Y = \frac{a X + b}{c}\] , where *a*, *b*, *c* are constants such that *ac* < 0, then

\[\sigma_Y = \frac{a}{c} \sigma_X\]

\[\sigma_Y = - \frac{a}{c} \sigma_X\]

\[\sigma_Y = \frac{a}{c} \sigma_X + b\]

none of these

If for a sample of size 60, we have the following information \[\sum_{} x_i^2 = 18000\] and \[ \sum_{} x_i = 960 \] , then the variance is

6.63

16

22

44

Let *a*, *b*, *c*, *d*, *e *be the observations with mean *m* and standard deviation *s*. The standard deviation of the observations *a* + *k*, *b* + *k*, *c* + *k*, *d* + *k*, *e* + *k* is

*s*

*ks*

*s* + *k*

\[\frac{s}{k}\]

Consider the first 10 positive integers. If we multiply each number by −1 and then add 1 to each number, the variance of the numbers so obtained is

8.25

6.5

3.87

2.87

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is

6.5

2.87

3.87

8.25

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is

50,000

250,000

252500

255000

Let *x*_{1}, *x*_{2}, ..., *x _{n}* be

*n*observations. Let \[y_i = a x_i + b\] for

*i*= 1, 2, 3, ...,

*n*, where

*a*and

*b*are constants. If the mean of \[x_i 's\] is 48 and their standard deviation is 12, the mean of \[y_i 's\] is 55 and standard deviation of \[y_i 's\] is 15, the values of

*a*and

*b*are

*a* = 1.25, *b* = −5

*a* = −1.25, *b* = 5

*a* = 2.5, *b* = −5

*a* = 2.5, *b* = 5

The mean deviation for *n* observations \[x_1 , x_2 , . . . , x_n\] from their mean \[\bar{X} \] is given by

\[\sum^n_{i = 1} \left( x_i - X \right)\]

\[\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)\]

\[\sum^n_{i = 1} \left( x_i - X \right)^2\]

\[\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)^2\]

Let \[x_1 , x_2 , . . . , x_n\] be *n* observations and \[X\] be their arithmetic mean. The standard deviation is given by

\[\sum^n_{i = 1} \left( x_i - X \right)^2\]

\[\frac{1}{n}\sum^n_{i = 1}\left( x_i - X \right)^2\]

\[\sqrt{\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)^2}\]

\[\sqrt{\frac{1}{n} \sum^n_{i = 1} x_i^2 - X^2}\]

The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is

6

\[\sqrt{6}\]

\[\frac{52}{7}\]

\[\sqrt{\frac{52}{7}}\]

#### RD Sharma Mathematics Class 11

#### Textbook solutions for Class 11

## RD Sharma solutions for Class 11 Mathematics chapter 32 - Statistics

RD Sharma solutions for Class 11 Maths chapter 32 (Statistics) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Class 11 solutions in a manner that help students grasp basic concepts better and faster.

Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 11 Mathematics chapter 32 Statistics are Standard Deviation - by Short Cut Method, Measures of Dispersion - Quartile Deviation, Central Tendency - Mode, Central Tendency - Mean, Statistics Concept, Comparison of Two Frequency Distributions with Same Mean, Introduction of Analysis of Frequency Distributions, Shortcut Method to Find Variance and Standard Deviation, Standard Deviation of a Continuous Frequency Distribution, Standard Deviation of a Discrete Frequency Distribution, Standard Deviation, Introduction of Variance and Standard Deviation, Mean Deviation, Concept of Range, Measures of Dispersion, Central Tendency - Median.

Using RD Sharma Class 11 solutions Statistics exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer RD Sharma Textbook Solutions to score more in exam.

Get the free view of chapter 32 Statistics Class 11 extra questions for Maths and can use shaalaa.com to keep it handy for your exam preparation