
01. Real Numbers
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Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9


02. Polynomials
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Lecture2.1

Lecture2.2

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Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03. Linear Equation
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Lecture3.1

Lecture3.2

Lecture3.3

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Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9


04. Quadratic Equation
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Lecture4.1

Lecture4.2

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Lecture4.4

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Lecture4.6

Lecture4.7

Lecture4.8


05. Arithmetic Progressions
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Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06. Some Applications of Trigonometry
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Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7


07. Coordinate Geometry
17
Lecture7.1

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Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

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Lecture7.11

Lecture7.12

Lecture7.13

Lecture7.14

Lecture7.15

Lecture7.16

Lecture7.17


08. Triangles
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Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

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Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13

Lecture8.14

Lecture8.15


09. Circles
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Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8


10. Areas Related to Circles
10
Lecture10.1

Lecture10.2

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Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9

Lecture10.10


11. Introduction to Trigonometry
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Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12. Surface Areas and Volumes
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Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. Statistics
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Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

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Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

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Lecture13.11

Lecture13.12


14. Probability
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Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Statistics
Q.1 A survey was conducted by a group of students as a part of their environment awareness
programmer, 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.
Which method did you use for finding the mean, and why ?
Sol. Calculation of Mean
Hence, mean x¯=1nΣfixi
= 120×162=8.1 plants
We have used direct method because numerical values of xi and fi are small.
Q.2 Consider the following distribution of daily wages of 50 workers of a factory.
Find the mean daily wages of the workers of the factory by using an appropriate method.
Sol. Let the assumed mean, A =150; classinterval, h = 20
so ui=xi−Ah=xi−15020
We construct the table :
x¯=A+hΣfiuiΣfi=150+20×−1250 = 150 – 4.8 = 145.2
Hence, mean = Rs 145.20 .
Q.3 The following distribution shows the daily pocket allowance of children locality mean. The mean pocket allowance is Rs 18. Find the missing frequency f.
Sol. Let the assumed mean, A = 16, class interval h = 2
So ui=xi−Ah=xi−162
We have, x¯ = 18, A = 16 and h = 2
Therefore, x¯=A+h(1NΣfiui)
⇒18=16+2(2f+24f+44)
⇒2=2(2f+24f+44)
⇒f+44=2f+24
⇒2f – f = 44 – 24
⇒f=20
Hence, the missing frequency is 20.
Q.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.
.
Sol. Let the assumed mean, A = 75.5, class interval
h = 3, so ui=xi−Ah=xi−75.53
x¯=A+hΣfiuiΣfi=75.5+3×430
= 75.5 + 0.4 = 75.9
Hence, the mean heart beats per minute for these women is 75.9.
Q.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.
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?
Sol. Here, the class intervals are formed by the exclusive method. If we make the series an inclusive one the midvalues remain same. So there is no need to convert the series.
Let the assumed mean be A = 60 and h = 3
so ui=xi−Ah=xi−603
Calculation of Mean
Therefore x¯=A+hΣfiuiΣfi=60+3×−375400
= 60 – 2.8125 = 57.1875 = 57.19(nearly)
Hence, average number of mangoes per box = 57.19 (nearly)
Q.6 The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
Sol. Let the assumed mean, A = 225, class interval, h = 50
so ui=xi−Ah=xi−22550
We construct the following table :
x¯=A+hΣfiuiΣfi
= 225 +50×−725
= 225 – 14 = 211
Hence, the mean daily expenditure of food is Rs 211.
Q.7 To find 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 :
Find the mean concentration of SO2 in the air.
Sol. Calculation of mean by direct method
Hence, x¯=∑fixi∑fi = 2.9630 = 0.099 ppm
Q.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.
Sol. Here, the class size varies, and xi ‘s are small. Let us apply the direct method here.
Therefore, mean, x¯=ΣfixiΣfi = 49940 = 12.475
Hence, mean = 12.48 days
Q.9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Sol. Let the assumed mean be A = 70 and h = 10
so ui=xi−7010
Therefore, x¯=A+h×ΣfiuiΣfi
= 70 + 10 ×−235
= 70 – 0.57 = 69.43
Thus, mean literacy rate is 69.43% .
Q.1 The following table shows the ages of the patients admitted in a hospital during a year :
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Sol. The class 35 – 45 has the maximum frequency, therefore, this is the modal class.
Here ℓ = 35, h = 10, f1 = 23, f0 = 21, f2 =14
Now, let us substitute these values in the formula
Mode =ℓ+(f1−f02f1−f0−f2)×h
= 35 + 23−212×23−21−14×10
=35+246−21−14×10
= 35 + 211×10
= 35 + 1.8 = 36.8
Calculation of Mean
Let assumed mean, A = 30, class interval h = 10
so ui=xi−Ah=xi−3010
Therefore, x¯=A+h×ΣfiuiΣfi = 30 + 10 ×4380
= 30 + 5.37 = 35.37
Hence, Mode = 36.8 years, Means = 35.37 years.
Maximum number of patients admitted in the hospital are of age group 36.8 years (approx.) while on an
average the age of a patient admitted to the hospital is 35.37 years.
Q.2 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Determine the modal lifetimes of the components.
Sol. The class 60 – 80 has the maximum frequency, therefore, this is the modal class.
Here ℓ = 60 , h = 20, f1 = 61, f0 = 52 and f2 = 38 .
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
=60+61−522×61−52−38×20
= 60+(61−52122−52−38)×20
= 60+932×20= 60 + 5.625 = 65.625
Thus, the modal lifetimes of the components is 65.625 hours.
Q.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 :
Sol. The class 1500 – 2000 has the maximum frequency,therefore, this is the modal class.
Here ℓ = 1500, h = 500, f1=40,f0=24 and f2=33
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
=1500+(40−242×40−24−33)×500
= 1500+40−2480−24−33×500
= 1500+1623×500
= 1500 + 347.83 = 1847.83
Therefore, modal monthly expenditure = Rs. 1847.83
Let the assumed mean be A = 3250 and h = 500.
We have, N = 200, A = 3250, h = 500 and Σfiui=−235
Therefore, mean = A+h(1NΣfiui)
= 3250 + 500 ×−235200
= 3250 – 587.5 = 2662.5
Hence, the average expenditure is Rs 2662.50.
Q.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.
Sol. The class 30 – 35 has the maximum frequency, therefore, this is the modal class.
Here ℓ = 30, h = 5, f1 =10, f0 = 9 and f2 =3
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
=30+(10−92×10−9−3)×5
= 30 + 10−920−9−3×5
= 30 +18×5
= 30 + 0.625
= 30.625 = 30.6(approx.)
Calculation of mean :
x¯=ΣfiuiΣfi
= 1022.535 = 29.2
Therefore, mean = 29.2
Thus, most states/U.T. have a student teacher ratio of 30.6 and on an average, the ration is 29.2.
Q.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.
Sol. Calculation of median
First, we prepare the following table to compute the median :
We have n = 68 ⇒n2=682=34
The cumulative frequency just greater than n2, is 42 and the corresponding class is 125 – 145.
Thus, 125 – 145is the median class such that n2 = 34, ℓ = 125, c.f. = 22, f = 20, h = 20 .
Substituting these values in the formula
Median = ℓ+(n2−cff)×h,
Median = 125 + (34−2220)×20 = 125 – 12 = 137 units
Calculation of mean
Let the assumed mean, A = 135,class interval, h = 20
so ui=xi−Ah=xi−13520
x¯=A+h×ΣfiuiΣfi
= 135 + 20 ×768
135 + 2.05 = 137. 05
Therefore, Mean = 137.05 units
Calculation of mode
The class 125 – 145 has the maximum frequency,therefore, this is the modal class.
Here ℓ =125, h =20, f1 =20, f0 = 13 and f2 =14
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
=125+20−1340−13−14×20
=125+713×20
= 125 + 10.76 = 135.76
Therefore, mode = 135.76 units
Clearly, the three measures are approximately the same in this case.
Q.2 If the median of the distribution given below is 28.5, find the values of x and y.
Sol. Here it is given that median is 28.5 and n = Σfi = 60
We now prepare the following cumulative frequency table :
Here n = 60 ⇒n2=30
Since the median is given to be 28.5, thus the median class is 20 – 30.
Therefore, ℓ = 20, h = 10, f = 20 and cf = 5 + x
Therefore, median = ℓ+(n2−cff)×h
⇒ 28.5 = 20+30−(5+x)20×10
⇒ 28.5 = 20 + 25−x2
⇒ 57 = 40 + 25 – x
⇒ x = 65 – 57 = 8
⇒ Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60 [As x = 8]
⇒ y = 60 – 53 = 7
Hence, x = 8, y = 7
Q.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 year.
Sol. We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median .
Here n = Σfi=100⇒n2=50
We see that the cumulative frequency just greater than n2, i.e., 50 is 78 and the corresponding class is 35 – 40. So, 35 – 40 is the median class.
Therefore, n2=50,ℓ=35, c.f = 45 , f = 33, h = 5
Now, let us substitute these values in the formula
Median = ℓ+(n2−cff)×h
= 35+50−4533×5 =35+533×5
= 35 + 0.76 = 35.76
Hence, the median age = 35.76 years .
Q.4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Find the median length of the leaves.
Sol. Here the frequency table is given in the inclusive form. So, we first convert it into exclusive form by subtracting and adding h2 to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and the upper limit of the previous class. Converting the given table into exclusive form and preparing the cumulative frequency table, we get
We have n = 40 ⇒n2=20
The cumulative frequency just greater than n2 is 29 and the corresponding class is 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
Here n2=20,ℓ = 144.5, h = 9, f = 12, cf =17
Substituting these values in the formula
Median = ℓ+(n2−cff)×h, we have
Median = 144.5+(20−1712)×9
= 144.5 +312×9
= 144.5 + 2.25
= 146.75 mm
Q.5 The following table gives the distribution of the life time of 400 neon lamps:
Find the median life time of a lamp.
Sol. First, we prepare the following table to compute the median:
We have n = 400 ⇒n2=200
The cumulative frequency just greater than n2 is 216 and the corresponding class is 3000 – 3500.
Thus, 3000 – 3500 is the median class such that n2=200,ℓ = 3000, cf = 130, f = 86 and h = 500.
Substituting these values in the formula
Median = ℓ+(n2−cff)×h, we have
Median = 3000+(200−13086)×500
= 3000+7086×500
= 3000 + 406.98 = 3406.98
Hence, median life = 3406.98 hours
Q.6 100 surnames were randomly picked up from a local telephone directory and the frequency
distribution of the number of letters in English alphabets 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.
Sol. Calculation of median
First, we prepare the following table to compute the median
We have n = 100 ⇒n2=50
The cumulative frequency just greater than n2 is 76 and the corresponding class is 7 – 10 . Thus 7 – 10 is the median class such that
n2=50,ℓ = 7, cf = 36, f = 40 and h = 3
Substitute these values in the formula
Median = ℓ+(n2−cff)×h, we have
Median = 7+(50−3640)×3
=7+1440×3
= 7 + 1.05 = 8.05
Calculation of mean
Therefore, Mean, x¯=ΣfixiΣfi=832100 = 8.32
Calculation of mode
The class 7 – 10 has the maximum frequency therefore, this is the modal class.
Here ℓ = 7 , h = 3, f1 = 40, f0 = 30 and f2 = 16
Now, let us substitute these values in the formula
Mode=ℓ+(f1−f02f1−f0−f2)×h
= 7+40−3080−30−16×3
= 7+1034×3 = 7 + 0.88 = 7.88
Hence, median = 8.05, mean = 8.32 and mode = 7.88
Q.1 The following distribution gives the daily income of 50 workers if a factory.
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Sol. Converting the given distribution to a less than type cumulative frequency distribution, we get
Let us now plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and join them by a free hand smooth curve.
The curve thus obtained is less than type ogive.
Q.2 During the medical checkup 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.
Sol. Here the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective classintervals. To represent the data in the table graphically, we mark the upper limits on the class intervals on xaxis and
their corresponding cumulative frequencies on the yaxis, choosing a convenient scale. Let us now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them by a free hand smooth curve.
The curve thus obtained is less than type ogive To locate n2=352= 17.5 on the yaxis. From this point draw a line parallel to the xaxis cutting the curve at a point. From this point, draw a perpendicular to the xaxis. The point of intersection of this perpendicular with the xaxis gives the median of the data. In this case it is 46.95. Let us make the following table in order to find mode by using the formula
The class 46 – 48 has the maximum frequency, therefore, this is modal class
Here, ℓ = 46, h = 2, f1 = 14, f0 = 5 and f2 = 4
Now, Mode = ℓ+(f1−f02f1−f0−f2)×h
= 46+14−528−5−4×2 = 46+919×2
= 46 + 0.95 = 46.95
Thus, mode is verified.
Q.3 The following tables 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.
Sol. Converting the given distribution to a more than type distribution, we get
.
Now, draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on
this graph paper and join them by a free hand smooth curve.
The curve thus obtained is more than type ogive.
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