Statistics Solution of TS & AP Board Class 10 Mathematics

Statistics Solution of TS & AP Board Class 10 Mathematics

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Exercise 14.1

Question 1.

A survey was conducted by a group of students as a part of their environment awareness programmed, 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.

Answer:

Let’s find mean of the data using direct method.

First, construct a table for ease of calculation.

Mean is given by

⇒ Mean = 162/20

⇒ Mean = 8.1

Thus, mean number of plants per house is 8.1.

Question 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.

Answer:

Let’s find mean of the data using assumed mean method.

We are using assumed mean method to avoid miscalculation and false answer, as x_i’s are large in this question.

First, construct a table for ease of calculation.

Mean is given by

Where A = assumed mean

⇒ 

⇒ Mean = 325 – 12 = 313

Thus, mean daily wages of the workers is Rs. 313.

Question 3.

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 18. Find the missing frequency f.

Answer:

To find f, we’ll need to find mean pocket allowance by direct method and equate it to the given mean pocket allowance, 18.

So, let’s construct a table finding midpoints and stating frequencies.

Mean is given by

⇒  [given, mean = 18]

⇒ 18(44 + f) = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 20f – 18f = 792 – 752

⇒ 2f = 40

⇒ f = 40/2

⇒ f = 20

Thus, the missing frequency f is 20.

Question 4.

Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarized as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Answer:

We shall use assumed mean method to find mean heart beats per minute, as this data has large values of x_i’s and has a risk of miscalculation. But by using assumed mean method, we’ll be able to solve it with ease and more accuracy.

Mean is given by

⇒ 

⇒ 

⇒ 

⇒ Mean = 75.9

Thus, mean heart beats per minute of these women is 75.9.

Question 5.

In a retail market, fruit vendors were selling oranges kept in packing basket. These baskets contained varying number of oranges. The following was distribution of oranges.

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?

Answer:

We shall use assumed mean method to find mean number of oranges.

So, while converting data into exclusive type won’t change any calculation in finding mean, we are however converting the inclusive data type to exclusive data type for better representation.

Mean is given by

⇒ 

⇒ Mean = 22 + 0.3125

⇒ Mean = 22.3125

Thus, mean number of oranges is 22.31.

We have used assumed mean method here, because of larger data values of f_i’s. Assumed mean method gives result to accuracy and ease when data is large.

Question 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.

Answer:

Let’s find mean of the data using assumed mean method for ease of calculation.

Mean is given by

⇒ 

⇒ Mean = 225 – 14

⇒ Mean = 211

Thus, mean daily expenditure on food is 211.

Question 7.

To find out the concentration of  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  in the air.

Answer:

Let’s find mean of the data using direct method as data given have smaller values.

First, construct a table for ease of calculation.

Mean is given by

⇒ Mean = 2.96/30

⇒ Mean = 0.0987

Or approximately, mean = 0.099

Thus, mean concentration of SO₂ in the air is 0.099 ppm.

Question 8.

A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term

Answer:

Here, we can find mean by either direct or assumed mean method. Let’s find mean of the data using direct method as data given have smaller values.

First, construct a table for ease of calculation.

Mean is given by

⇒ Mean = 1961/40

⇒ Mean = 49.025

Or approximately, mean = 49

Thus, mean number of days a student was present is 49.

Question 9.

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

Let’s find mean of the data using direct method as data given have smaller values.

First, construct a table for ease of calculation.

Mean is given by

⇒ Mean = 2430/35

⇒ Mean = 69.429

Or approximately, mean = 69.43

Thus, mean literacy rate is 69.43%.

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Exercise 14.2

Question 1.

The following table shows the ages of the patients admitted in a hospital during a year :

Find the made and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer:

For mode:

Here, highest frequency is 23.

So, the modal class = 35-45

Mode is given by

Where,

L = Lower class limit of the modal class = 35

h = class interval of the modal class = 10

f₁ = frequency of the modal class = 23

f₀ = frequency of the class preceding the modal class = 21

f₂ = frequency of the class succeeding the modal class = 14

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 35 + 20/11

⇒ Mode = 36.8

For mean:

Let’s draw a table showing midpoints and frequencies.

Mean is given by

⇒ Mean = 2830/80

⇒ Mean = 35.37

Thus, mode is 36.8 and mean is 35.37.

The mode shows maximum number of patients admitted in the hospital is of age 36.8 years approximately and mean shows that on an average, number of patients admitted is of age 35.37 years.

Question 2.

The following data gives the information on the observed life times (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Answer:

For mode:

Here, highest frequency is 61.

So, the modal class = 60-80

Mode is given by

Where,

L = Lower class limit of the modal class = 60

h = class interval of the modal class = 20

f₁ = frequency of the modal class = 61

f₀ = frequency of the class preceding the modal class = 52

f₂ = frequency of the class succeeding the modal class = 38

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 60 + 180/32

⇒ Mode = 60 + 5.625

⇒ Mode = 65.625

Thus, modal lifetime of the components is 65.625 years.

Question 3.

The following data gives the distribution of total monthly household expenditure of 200 families of Gummandidala village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Answer:

For mode:

Here, highest frequency is 40.

So, the modal class = 1500-2000

Mode is given by

Where,

L = Lower class limit of the modal class = 1500

h = class interval of the modal class = 500

f₁ = frequency of the modal class = 40

f₀ = frequency of the class preceding the modal class = 24

f₂ = frequency of the class succeeding the modal class = 33

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 1500 + 347.8

⇒ Mode = 1847.8

For mean:

Using assumed-mean method, as values of the data are large.

Let’s draw a table showing midpoints and frequencies.

Mean is given by

⇒ 

⇒ Mean = 2750 – 87.5

⇒ Mean = 2662.5

Thus, mode is 1847.8 and mean is 2662.5.

Question 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.

Answer:

For mode:

Here, highest frequency is 10.

So, the modal class = 30-35

Mode is given by

Where,

L = Lower class limit of the modal class = 30

h = class interval of the modal class = 5

f₁ = frequency of the modal class = 10

f₀ = frequency of the class preceding the modal class = 9

f₂ = frequency of the class succeeding the modal class = 3

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 30 + 0.625

⇒ Mode = 30.625

Or approximately, mode = 30.6

For mean:

Let’s solve by using direct method as values are small.

Let’s draw a table showing midpoints and frequencies.

Mean is given by

⇒ Mean = 1022.5/35

⇒ Mean = 29.2

Thus, mode is 30.6 and mean is 29.2.

Mode implies that most states have student teacher ratio of 30.6 and mean implies that on an average, student teacher ratio in most states is found to be 29.2.

Question 5.

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Answer:

For mode:

Here, highest frequency is 18.

So, the modal class = 4000-5000

Mode is given by

Where,

L = Lower class limit of the modal class = 4000

h = class interval of the modal class = 1000

f₁ = frequency of the modal class = 18

f₀ = frequency of the class preceding the modal class = 4

f₂ = frequency of the class succeeding the modal class = 9

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 4000 + 608.7

⇒ Mode = 4608.7

Thus, the mode of the data is 4608.7.

Question 6.

A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarized this in the table given below.

Find the mode of the data.

Answer:

For mode:

Here, highest frequency is 20.

So, the modal class = 40-50

Mode is given by

Where,

L = Lower class limit of the modal class = 40

h = class interval of the modal class = 10

f₁ = frequency of the modal class = 20

f₀ = frequency of the class preceding the modal class = 12

f₂ = frequency of the class succeeding the modal class = 11

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 40 + 4.7

⇒ Mode = 44.7

Thus, the mode of the data is 44.7.

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Exercise 14.3

Question 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.

Answer:

Let’s draw a table showing midpoints, frequencies and cumulative frequency.

For median:

We have, total frequency, N = 68

N/2 = 68/2 = 34

Observe, cf = 42 is just greater than 34.

Thus, median class = 125-145

Median is given by

Where,

L = Lower class limit of median class = 125

N/2 = 34

cf = cumulative frequency of the class preceding median class = 22

f = frequency of the median class = 20

h = class interval of the median class = 20

Substituting these values in the formula of median, we get

⇒ Median = 125 + 12

⇒ Median = 137

For mean:

Mean is given by

⇒ 

⇒ Mean = 135 + 2.06

⇒ Mean = 137.06

For mode:

Here, highest frequency is 20.

So, the modal class = 125-145

Mode is given by

Where,

L = Lower class limit of the modal class = 125

h = class interval of the modal class = 20

f₁ = frequency of the modal class = 20

f₀ = frequency of the class preceding the modal class = 13

f₂ = frequency of the class succeeding the modal class = 14

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 125 + 10.76

⇒ Mode = 135.76

Thus, median is 137, mean is 137.06 and mode is 135.76.

Median and mean, both gives us the average value of a data, with just the difference that mean can give us quantitative measure only but median can give us both quantitative as well as qualitative measure of a data. And that is why, mean and median have come out to be very close in this question.

Mode gives the value that appears most in a given data. Thus, the maximum monthly consumption of electricity is 135.76.

Question 2.

If the median of 60 observations, given below is 28.5, find the values of x and y.

Answer:

For median:

Since, given that: Median = 28.5

We know, median class = 20-30

Median is given by

Where,

L = Lower class limit of median class = 20

N/2 = 60/2 = 30 [∵, given: total observation = 60]

cf = cumulative frequency of the class preceding median class = 5 + x

f = frequency of the median class = 20

h = class interval of the median class = 10

Substituting these values in the formula of median, we get

⇒ Median = 20 + (25 – x)/2

⇒ 28.5 = (40 + 25 – x)/2

⇒ 57 = 65 – x

⇒ x = 65 – 57 = 8

Recall, N = 45 + x + y = 60 [given]

And x = 8

⇒ 45 + 8 + y = 60

⇒ 53 + y = 60

⇒ y = 60 – 53

⇒ y = 7

Thus, x = 8 and y = 7.

Question 3.

A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.]

Answer:

For median:

We have, total frequency, N = 534

N/2 = 534/2 = 267

Observe, cf = 336 is just greater than 267.

Thus, median class = 45-50

Median is given by

Where,

L = Lower class limit of median class = 45

N/2 = 267

cf = cumulative frequency of the class preceding median class = 244

f = frequency of the median class = 92

h = class interval of the median class = 5

Substituting these values in the formula of median, we get

⇒ 

⇒ Median = 45 + 115/92

⇒ Median = 45 + 1.25

⇒ Median = 46.25

Thus, the median is 46.25.

Question 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. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5.)

Answer:

We’ll convert this inclusive data into exclusive data type.

For median:

We have, total frequency, N = 40

N/2 = 40/2 = 20

Observe, cf = 29 is just greater than 20.

Thus, median class = 144.5-153.5

Median is given by

Where,

L = Lower class limit of median class = 144.5

N/2 = 20

cf = cumulative frequency of the class preceding median class = 17

f = frequency of the median class = 12

h = class interval of the median class = 9

Substituting these values in the formula of median, we get

⇒ 

⇒ Median = 144.5 + 9/4

⇒ Median = 144.5 + 2.25

⇒ Median = 146.75

Thus, the median is 146.75 mm.

Question 5.

The following table gives the distribution of the life-time of 400 neon lamps

Find the median life time of a lamp.

Answer:

For median:

We have, total frequency, N = 400

N/2 = 400/2 = 200

Observe, cf = 216 is just greater than 200.

Thus, median class = 3000-3500

Median is given by

Where,

L = Lower class limit of median class = 3000

N/2 = 200

cf = cumulative frequency of the class preceding median class = 130

f = frequency of the median class = 86

h = class interval of the median class = 500

Substituting these values in the formula of median, we get

⇒ 

⇒ Median = 3000 + 406.98

⇒ Median = 3406.98

Thus, the median is 3406.98 hours.

Question 6.

100 sumames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet 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.

Answer:

For median:

We have, total frequency, N = 100

N/2 = 100/2 = 50

Observe, cf = 76 is just greater than 50.

Thus, median class = 7-10

Median is given by

Where,

L = Lower class limit of median class = 7

N/2 = 50

cf = cumulative frequency of the class preceding median class = 36

f = frequency of the median class = 40

h = class interval of the median class = 3

Substituting these values in the formula of median, we get

⇒ 

⇒ Median = 7 + 1.05

⇒ Median = 8.05

For mean:

Mean is given by

⇒  [Refer the values from the table given above]

⇒ Mean = 8.32

For mode:

Here, highest frequency is 40.

So, the modal class = 7-10

Mode is given by

Where,

L = Lower class limit of the modal class = 7

h = class interval of the modal class = 3

f₁ = frequency of the modal class = 40

f₀ = frequency of the class preceding the modal class = 30

f₂ = frequency of the class succeeding the modal class = 16

Substituting values in the formula of mode,

⇒ 

⇒ Mode = 7 + 0.88

⇒ Mode = 7.88

Thus, median is 8.05, mean is 8.32 and mode is 7.88.

Question 7.

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

For median:

We have, total frequency, N = 30

N/2 = 30/2 = 15

Observe, cf = 19 is just greater than 15.

Thus, median class = 55-60

Median is given by

Where,

L = Lower class limit of median class = 55

N/2 = 15

cf = cumulative frequency of the class preceding median class = 13

f = frequency of the median class = 6

h = class interval of the median class = 5

Substituting these values in the formula of median, we get

⇒ 

⇒ Median = 55 + 1.67

⇒ Median = 56.67

Thus, the median is 56.67 kg.

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Exercise 14.4

Question 1.

The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Answer:

Let’s make less than type cumulative frequency table:

Plot points (300,12), (350,26), (400, 34), (450, 40) and (500,50) on a graph.

Question 2.

During the medical check-up 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.

Answer:

Using points (38,0), (40,3), (42,5), (44,9), (46,14), (48,28), (50, 32) and (52,35), plot a graph.

To find median graphically,

We have total frequency, N = 35

⇒ N/2 = 35/2 = 17.5

Plot 17.5 on the y-axis and draw a horizontal line intersecting the ogive parallel to x-axis.

Observe, from the graph the vertical line parallel to y-axis touches y-axis at 46.5 (approx.).

Hence, median is 46.5.

Finding median by formula:

For median:

We have, total frequency, N = 35

N/2 = 35/2 = 17.5

Observe, cf = 28 is just greater than 17.5.

Thus, median class = 46-48

Median is given by

Where,

L = Lower class limit of median class = 46

N/2 = 17.5

cf = cumulative frequency of the class preceding median class = 14

f = frequency of the median class = 14

h = class interval of the median class = 2

Substituting these values in the formula of median, we get

⇒ 

⇒ Median = 46 + 0.5

⇒ Median = 46.5

Thus, median is 46.5.

Hence, verified.

Question 3.

the following table 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.

Answer:

Let’s make more than type cumulative frequency table:

Plot points (50,100), (55,98), (60, 90), (65, 78), (70, 54) and (75,16) on a graph.