Statistics Class 10th Mathematics Gujarat Board Solution

Class 10^th Mathematics Gujarat Board Solution

Exercise 15.1

1. Find the mean of the following frequency
2. Find the mean wage of 200 workers of a factory where wages are classified as…
3. Marks obtained by 140 students of class X out of 50 in mathematics are given in…
4. Find the mean of the following frequency distribution by step-deviation…
5. Find the mean for the following frequency
6. A survey conducted by a student of B.B.A. for daily income of 600 families is…
7. The number of shares held by a person of various companies are as follows. Find…
8. The mean of the following frequency distribution of 100 observations is 148.…
9. The table below gives the percentage of girls in higher secondary science…
10. The following distribution shows the number of out door patients in 64…

Exercise 15.2

1. Find the mode for the following frequency
2. The data obtained for 100 shops for their daily profit per shop are as…
3. Daily wages of 90 employees of a factory are as follows:Daily wages (in…
4. Find the mode for the following
5. Find the mode for the following
6. The following data gives the information of life of 200 electric bulbs (in…

Exercise 15.3

1. Find the median for the following:Value of
2. Find the median for the following frequency
3. Find the median from following frequency
4. The following frequency distribution represents the deposits (in thousand…
5. The median of the following frequency distribution is 38. Find the value of a…
6. The median of 230 observations of the following frequency distribution is 46.…
7. The following table gives the frequency distribution of marks scored by 50…

Exercise 15

1. In a retail market, a fruit vendor was selling apples kept in packed boxes.…
2. The daily expenditure of 50 hostel students are as follows:Daily expenditure…
3. The mean of the following frequency distribution of 200 observations is 332.…
4. Find the mode of the following frequency
5. Find the mode of the following
6. The mode of the following frequency distribution of 165 observations is 34.5.…
7. Find the mode of the following frequency
8. Find the median of the following frequency
9. The median of the following data is 525. Find the value of x and y, if the sum…
10. For some data, if Z = 25 and bar {x} = 25, then M = Select a proper option…
11. For some data Z — M = 2.5. If the mean of the data is 20, then Z = Select a…
12. If bar {x} — Z = 3 and bar {x} + Z = 45, then M = Select a proper option…
13. If Z = 24, bar {x} = 18, then M= Select a proper option (a), (b), (c) or…
14. If M = 15, bar {x} = 10, then Z= Select a proper option (a), (b), (c) or (d)…
15. (6) If M = 22, Z = 16, then bar {x} = Select a proper option (a), (b), (c)…
16. (7) If bar {x} = 21.44 and Z = 19.13, then M = Select a proper option (a),…
17. If M = 26, bar {x} =36, then Z = Select a proper option (a), (b), (c) or…
18. (9) The modal class of the frequency distribution given below
19. The cumulative frequency of class 20-30 of the frequency distribution given…
20. The median class of the frequency distribution given in (9) is …. Select a…

---

Exercise 15.1

Question 1.

Find the mean of the following frequency distribution:

Answer:

Since, this is a grouped frequency distribution, this is solved by creating a table and using certain formulae. To find mean, we need to create a table showing class intervals, midpoints, frequencies and product of those midpoints and frequencies.

Now, we have

∑f_i = 100 and ∑x_if_i = 14850

The formula of mean by direct method is given by,

⇒  [from the given information]

⇒ Mean = 148.5

Thus, mean is 148.5.

Question 2.

Find the mean wage of 200 workers of a factory where wages are classified as follows :

Answer:

This is grouped frequency distribution.

We shall represent them by creating a table having columns of class intervals, midpoints, frequencies, and product of these midpoints and frequencies.

So now, we have

∑f_i = 200 and ∑x_if_i = 64300

For grouped frequency distribution, mean is taken out from the formula:

⇒  [from the table]

⇒ 

⇒ Mean = 321.5

Thus, mean is 321.5.

Question 3.

Marks obtained by 140 students of class X out of 50 in mathematics are given in the following distribution. Find the mean by method of assumed mean method:

Answer:

We have grouped frequency distribution and we need to find mean by assumed-mean method.

Assumed-mean method is given such name because in this method, we actually assume a mean from x_I (usually a centre value) and the formula of mean is also based on that particular assumed mean.

By using assumed-mean method, we can avoid the risk of miscalculation. Also, by using assumed mean method, we’ll be able to solve the question with ease and more accuracy.

So, let us assume mean from the midpoints. Let assumed mean, A = 25 (a value quite centrally placed).

So now, we have

∑f_id_i = 120 and ∑f_i = 140.

And we have assumed mean as, A = 25

Mean is given by

⇒ 

⇒ Mean = 25 + 0.86

⇒ Mean = 25.86

Thus, mean is 25.86.

Question 4.

Find the mean of the following frequency distribution by step-deviation method:

Answer:

This is a grouped frequency distribution.

In step deviation, we need to mark the assumed mean in x_i as similar to in assumed-mean method but the formula differs. In step deviation, we calculate the deviation of each value from the assumed mean value. This brings more accuracy when values are large.

Here, class interval is given by

h = 10 (∵ 50 – 40 = 60 – 50 = … = 100 – 90 = 10)

And let us assume mean as A = 65, which is the closest value to the centre-most value.

We need to represent it in tabular form:

So now, we have

∑f_iu_i = 7 and ∑f_i = 52

And assumed mean, A = 65

Mean by step-deviation is given by

⇒ 

⇒ 

⇒ Mean = 65 + 1.346

⇒ Mean = 66.346

Thus, mean is 66.346.

Question 5.

Find the mean for the following frequency distribution:

Answer:

This is a grouped frequency distribution.

We can find the mean of the data using any of the methods. Since here we have small values of classes, we can find the mean using direct method easily.

We need to create a table using the values, representing class, midpoints, frequency and product of midpoints and frequencies.

Also, this is an inclusive type of data, which needs to be converted into exclusive type of data.

For conversion, we need to subtract 0.5 from the lower class of each interval and add 0.5 to upper class of each interval.

So now, we have

∑x_if_i = 3735 and ∑f_i = 200

Mean is given by

⇒ [Using values from the table]

⇒ Mean = 18.675

Thus, mean is 18.675.

Question 6.

A survey conducted by a student of B.B.A. for daily income of 600 families is as follows, find the mean income of a family:

Answer:

This is a grouped frequency distribution.

We can find the mean of this type of distribution by either direct method, assumed-mean method or step deviation method.

But let us try doing it by assumed-mean method since it assures accuracy of the answer when data given is large.

Also, notice given data is inclusive-type of data. We need to convert the inclusive type of data into exclusive type by subtracting 0.5 from lower limit of each class and adding 0.5 to upper limit of each class.

Let us assume mean by taking a value from x_i’s closest from the central value.

Let assumed mean, A = 549.5

So, we have

So now, we have

∑f_id_i = 18500 and ∑f_i = 600

Mean is given by

⇒ 

⇒ Mean = 549.5 + 30.83

⇒ Mean = 580.33

Thus, mean is 580.33.

Question 7.

The number of shares held by a person of various companies are as follows. Find the mean:

Answer:

This is a grouped frequency distribution.

Notice the data in ‘number of companies’, is small and the class in ‘Number of shares’ have equal intervals. So, we can use direct method for ease of calculation and can rely on the answer for accuracy.

So, create the table again as we need to represent midpoints and frequency. Re-creating the table ensures ease of calculation.

The calculation is done on a table and we can easily extract data from it.

We have,

So now, we have

∑x_if_i = 7000 and ∑f_i = 20

Using direct method,

Mean is given by

⇒  [Using values in the table]

⇒ Mean = 350

Thus, mean is 350.

Question 8.

The mean of the following frequency distribution of 100 observations is 148. Find the missing frequencies f₁ and f₂ :

Answer:

This is a grouped frequency distribution.

To find f₁ and f₂, we’ll need to find mean of the following distribution by direct method and equate it to the given mean of the following distribution, 148.

This given data is in inclusive type, and we need not convert it into exclusive since it won’t make a difference in the calculation.

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

So now, we have

∑x_if_i = 8869 + 124.5f₁ + 274.5f₂

And ∑f_i = 62 + f₁ + f₂

Mean is given by

⇒  [given, mean = 148 and using the values from the table]

⇒ 148 × (62 + f₁ + f₂) = 8869 + 124.5f₁ + 274.5f₂

⇒ 9176 + 148f₁ + 148f₂ = 8869 + 124.5f₁ + 274.5f₂

⇒ 274.5f₂ – 148f₂ + 124.5f₁ – 148f₁ = 9176 – 8869

⇒ 126.5f₂ – 23.5f₁ = 307

⇒ 

⇒ 

⇒ 253f₂ – 47f₁ = 307 × 2

⇒ 253f₂ – 47f₁ = 614 …(i)

Since, there are 100 observations.

⇒ Frequency = 100 [Frequencies depict total number of observation]

⇒ ∑f_i = 100

⇒ 62 + f₁ + f₂ = 100

⇒ f₁ + f₂ = 100 – 62 = 38

⇒ f₁ + f₂ = 38 …(ii)

Multiplying 47 by equation (ii), we get

f₁ + f₂ = 38 [× 47

⇒ 47f₁ + 47f₂ = 1786 …(iii)

Solving equations (i) and (iii), we get

253f₂ – 47f₁ = 614

47f₂ + 47f₁ = 1786

300f₂ + 0 = 2400

⇒ 300f₂ = 2400

⇒ 

⇒ f₂ = 8

Putting f₂ = 8 in equation (ii), we get

f₁ + 8 = 38

⇒ f₁ = 38 – 8

⇒ f₁ = 30

Thus, the missing frequencies are f₁ = 30 and f₂ = 8.

Question 9.

The table below gives the percentage of girls in higher secondary science stream of rural areas of various states of India. Find the mean percentage of girls by step-deviation method:

Answer:

This is a grouped frequency distribution.

In step deviation, we need to mark the assumed mean in x_i. Step deviation also calculates deviation of each x_i values from the assumed mean using a formula.

The assumed mean is the value from x_i closest to the central value.

So, assumed mean is

A = 50

Here, class interval is given by

h = 10 (∵ 25 – 15 = 35 – 25 = … = 85 – 75 = 10)

We need to represent it in tabular form:

So now, we have

∑f_iu_i = -29 and ∑f_i = 35.

Also, assumed mean, A = 50 and class interval, h = 10.

Mean by step-deviation is given by

⇒  [using the values from the table]

⇒ 

⇒ Mean = 50 – 8.29

⇒ Mean = 41.71

Thus, mean is 41.71.

Question 10.

The following distribution shows the number of out door patients in 64 hospitals as follows. If the mean is 18, find the missing frequencies f₁ and f₂ :

Answer:

This is a grouped frequency distribution.

To find f₁ and f₂, we’ll need to find mean of the following distribution by direct method and equate it to the given mean of the following distribution, 18.

This given data is exclusive, grouped frequency distribution.

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

So now, we have

∑x_if_i = 608 + 16f₁ + 20f₂

And ∑f_i = 35 + f₁ + f₂

Mean is given by

⇒  [given, mean = 18 and using the values from the table]

⇒ 18 × (35 + f₁ + f₂) = 608 + 16f₁ + 20f₂

⇒ 630 + 18f₁ + 18f₂ = 608 + 16f₁ + 20f₂

⇒ 20f₂ – 18f₂ + 16f₁ – 18f₁ = 630 – 608

⇒ 2f₂ – 2f₁ = 22

⇒ 2(f₂ – f₁) = 22

⇒ f₂ – f₁ = 11 …(i)

Since, there are 64 hospitals.

⇒ Number of hospitals = 64

⇒ Frequency = 64

⇒ ∑f_i = 64

⇒ 35 + f₁ + f₂ = 64

⇒ f₁ + f₂ = 64 – 35

⇒ f₁ + f₂ = 29 …(ii)

Solving equations (i) and (ii), we get

f₂ – f₁ = 11

f₂ + f₁ = 29

2f₂ + 0 = 40

⇒ 2f₂ = 40

⇒ 

⇒ f₂ = 20

Putting f₂ = 20 in equation (ii), we get

f₁ + 20 = 29

⇒ f₁ = 29 – 20

⇒ f₁ = 9

Thus, the missing frequencies are f₁ = 9 and f₂ = 20.

---

Exercise 15.2

Question 1.

Find the mode for the following frequency distribution:

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 15 and the class corresponding to this frequency is 20 – 24.

So, this implies that,

Modal class = 20 – 24

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 20

f₀ = frequency of the class preceding the modal class = 7

f₁ = frequency of the modal class = 15

f₂ = frequency of the class succeeding the modal class = 1

c = size of class interval (the class intervals are same) = 4

∴ Substituting the values l = 20, f₀ = 7, f₁ = 15, f₂ = 1 and c = 4 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ 

⇒ Mode = 20 + 1.45

⇒ Mode = 21.45

Thus, the mode is 21.45.

Question 2.

The data obtained for 100 shops for their daily profit per shop are as follows:


Find the modal profit per shop.

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 27 and the class corresponding to this frequency is 200 – 300.

So, this implies that,

Modal class = 200 – 300

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 200

f₀ = frequency of the class preceding the modal class = 18

f₁ = frequency of the modal class = 27

f₂ = frequency of the class succeeding the modal class = 20

c = size of class interval (the class intervals are same) = 100

∴ Substituting the values l = 200, f₀ = 18, f₁ = 27, f₂ = 20 and c = 100 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ 

⇒ Mode = 200 + 56.25

⇒ Mode = 256.25

Thus, the modal profit per shop is Rs. 256.25.

Question 3.

Daily wages of 90 employees of a factory are as follows:


Find the modal wage of an employee.

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 33 and the class corresponding to this frequency is 550 – 650.

So, this implies that,

Modal class = 550 – 650

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 550

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

f₁ = frequency of the modal class = 33

f₂ = frequency of the class succeeding the modal class = 17

c = size of class interval (the class intervals are same) = 100

∴ Substituting the values l = 550, f₀ = 12, f₁ = 33, f₂ = 17 and c = 100 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ Mode = 550 + 56.76

⇒ Mode = 606.76

Thus, the modal wage of an employee is Rs. 606.76.

Question 4.

Find the mode for the following data:

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 82 and the class corresponding to this frequency is 28 – 35.

So, this implies that,

Modal class = 28 – 35

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 28

f₀ = frequency of the class preceding the modal class = 42

f₁ = frequency of the modal class = 82

f₂ = frequency of the class succeeding the modal class = 71

c = size of class interval (the class intervals are same) = 7

∴ Substituting the values l = 28, f₀ = 42, f₁ = 82, f₂ = 71 and c = 7 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ Mode = 28 + 5.49

⇒ Mode = 33.49

Thus, the mode is 33.49

Question 5.

Find the mode for the following data:

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 31 and the class corresponding to this frequency is 80 – 100.

So, this implies that,

Modal class = 80 – 100

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 80

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

f₁ = frequency of the modal class = 31

f₂ = frequency of the class succeeding the modal class = 27

c = size of class interval (the class intervals are same) = 20

∴ Substituting the values l = 80, f₀ = 21, f₁ = 31, f₂ = 27 and c = 20 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ 

⇒ Mode = 80 + 14.29

⇒ Mode = 94.29

Thus, the mode is 94.29

Question 6.

The following data gives the information of life of 200 electric bulbs (in hours) as follows:


Find the modal life of the electric bulbs.

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 82 and the class corresponding to this frequency is 80 – 100.

So, this implies that,

Modal class = 80 – 100

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 80

f₀ = frequency of the class preceding the modal class = 42

f₁ = frequency of the modal class = 82

f₂ = frequency of the class succeeding the modal class = 71

c = size of class interval (the class intervals are same) = 20

∴ Substituting the values l = 80, f₀ = 42, f₁ = 82, f₂ = 71 and c = 20 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ Mode = 80 + 15.69

⇒ Mode = 95.69

Thus, the modal life of electric bulbs is 94.29 hours.

---

Exercise 15.3

Question 1.

Find the median for the following:

Answer:

Here, we have ungrouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 100

Since, n (=100) is even then the median will be the average of  and  observations.

We have

⇒ 

And

⇒ 

⇒ 

Note that, in the above table:

50^th observation = 15 &

51^st observation = 16

[∵ 50 and 51 lies between 50 and 70 in the cumulative frequency column]

Taking their average, we get

⇒ 

⇒ 

⇒ Median = 15.5

Thus, the median is 15.5

Question 2.

Find the median for the following frequency distribution:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 60

Now, we just need to find the value of n/2. So,

⇒ 

Now, look up for a value in the cumulative frequency just greater than 30.

We have, 37.

Corresponding to this value of cumulative frequency, we can say that median class is 12 – 16.

That is,

Median class = 12 – 16

∴ we have almost everything we require to calculate median.

Median is given by,

Where,

l = lower limit of the median class = 12

n = Total number of observation (sum of frequencies) = 60

cf = cumulative frequency of the class preceding the median class = 25

f = frequency of the median class = 12

c = class size (class sizes are equal) = 4

Putting the values, l = 12, n/2 = 30, cf = 25, f = 12 and c = 4 in the given formula of median, we get

⇒ 

⇒ 

⇒ Median = 12 + 1.67

⇒ Median = 13.67

Thus, the median of the data is 13.67.

Question 3.

Find the median from following frequency distribution:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 400

Now, we just need to find the value of n/2. So,

⇒ 

Now, look up for a value in the cumulative frequency just greater than 200.

We have, 210.

Corresponding to this value of cumulative frequency, we can say that median class is 200 – 300.

That is,

Median class = 200 – 300

∴ we have almost everything we require to calculate median.

Median is given by,

Where,

l = lower limit of the median class = 200

n = Total number of observation (sum of frequencies) = 400

cf = cumulative frequency of the class preceding the median class = 126

f = frequency of the median class = 84

c = class size (class sizes are equal) = 100

Putting the values, l = 200, n/2 = 200, cf = 126, f = 84 and c = 100 in the given formula of median, we get

⇒ 

⇒ 

⇒ 

⇒ Median = 200 + 88.09

⇒ Median = 288.09

Thus, the median of the data is 288.09.

Question 4.

The following frequency distribution represents the deposits (in thousand rupees) and the number of depositors in a bank. Find the median of the data:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 2776

Now, we just need to find the value of n/2. So,

⇒ 

Now, look up for a value in the cumulative frequency just greater than 1388.

We have, 2316.

Corresponding to this value of cumulative frequency, we can say that median class is 10 – 20.

That is,

Median class = 10 – 20

∴ we have almost everything we require to calculate median.

Median is given by,

Where,

l = lower limit of the median class = 10

n = Total number of observation (sum of frequencies) = 2776

cf = cumulative frequency of the class preceding the median class = 1071

f = frequency of the median class = 1245

c = class size (class sizes are equal) = 10

Putting the values, l = 10, n/2 = 1388, cf = 1071, f = 1245 and c = 10 in the given formula of median, we get

⇒ 

⇒ 

⇒ 

⇒ Median = 10 + 2.55

⇒ Median = 12.55

Thus, the median of the data is Rs. 12.55 (in thousand).

Question 5.

The median of the following frequency distribution is 38. Find the value of a and b if the sum of frequences is 400:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 202 + a + b …(i)

But given is, sum of frequencies, ∑f = n = 400 …(ii)

⇒ 

⇒ 

Comparing equations (i) and (ii), we get

202 + a + b = 400

⇒ a + b = 400 – 202

⇒ a + b = 198 …(iii)

Also, given that, median = 38

Corresponding to this value of median, we can say that median lies in the class is 30 – 40. [∵ 38 lies between 30 – 40]

⇒ Median class = 30 – 40

Median is given by,

Where,

l = lower limit of the median class = 30

n = Total number of observation (sum of frequencies) = 400

cf = cumulative frequency of the class preceding the median class = 80

f = frequency of the median class = a

c = class size (class sizes are equal) = 10

Putting the values, l = 30, n/2 = 200, cf = 80, f = a and c = 10 in the given formula of median, we get

⇒  [∵ given that, median = 38]

⇒ 

⇒ 

⇒ 

⇒ 

⇒ a = 150

Substituting the value, a = 150 in equation (iii), we get

a + b = 198

⇒ 150 + b = 198

⇒ b = 198 – 150

⇒ b = 48

Thus, a = 150 and b = 48.

Question 6.

The median of 230 observations of the following frequency distribution is 46. Find a and b:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 150 + a + b …(i)

But given is, sum of frequencies, ∑f = n = 230 …(ii)

⇒ 

⇒ 

Comparing equations (i) and (ii), we get

150 + a + b = 230

⇒ a + b = 230 – 150

⇒ a + b = 80 …(iii)

Also, given that, median = 46

Corresponding to this value of median, we can say that median lies in the class is 40 – 50. [∵ 46 lies between 40 – 50]

⇒ Median class = 40 – 50

Median is given by,

Where,

l = lower limit of the median class = 40

n = Total number of observation (sum of frequencies) = 230

cf = cumulative frequency of the class preceding the median class = 42 + a

f = frequency of the median class = 65

c = class size (class sizes are equal) = 10

Putting the values, l = 40, n/2 = 115, cf = 42 + a, f = 65 and c = 10 in the given formula of median, we get

⇒  [∵ given that, median = 46]

⇒ 

⇒ 

⇒ 

⇒ 730 – 10a = 390

⇒ 10a = 730 – 390

⇒ 10a = 340

⇒ 

⇒ a = 34

Substituting the value, a = 34 in equation (iii), we get

a + b = 80

⇒ 34 + b = 80

⇒ b = 80 – 34

⇒ b = 46

Thus, a = 34 and b = 46.

Question 7.

The following table gives the frequency distribution of marks scored by 50 students of class X in mathematics examination of 80 marks. Find the median of the data:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 50

Now, we just need to find the value of n/2. So,

⇒ 

Now, look up for a value in the cumulative frequency just greater than 25.

We have, 31.

Corresponding to this value of cumulative frequency, we can say that median class is 30 – 40.

That is,

Median class = 30 – 40

∴ we have almost everything we require to calculate median.

Median is given by,

Where,

l = lower limit of the median class = 30

n = Total number of observation (sum of frequencies) = 50

cf = cumulative frequency of the class preceding the median class = 15

f = frequency of the median class = 16

c = class size (class sizes are equal) = 10

Putting the values, l = 30, n/2 = 25, cf = 15, f = 16 and c = 10 in the given formula of median, we get

⇒ 

⇒ 

⇒ 

⇒ Median = 30 + 6.25

⇒ Median = 36.25

Thus, the median of the data is 36.25.

---

Exercise 15

Question 1.

In a retail market, a fruit vendor was selling apples kept in packed boxes. These boxes contained varying number of apples. The following was the distribution of apples according to the number of boxes. Find the mean by the assumed mean number of apples kept in the box.

Answer:

We have grouped frequency distribution and we need to find mean by assumed-mean method.

Assumed-mean method is given such name because in this method, we actually assume a mean from x_I (usually a centre value) and the formula of mean is also based on that particular assumed mean.

By using assumed-mean method, we can avoid the risk of miscalculation. Also, by using assumed mean method, we’ll be able to solve the question with ease and more accuracy.

So, let us assume mean from the midpoints. Let assumed mean, A = 57.5 (a value quite centrally placed).

So now, we have

∑f_id_i = -165 and ∑f_i = 400.

And we have assumed mean as, A = 57.5

Mean is given by

⇒ 

⇒ Mean = 57.5 – 0.4125

⇒ Mean = 57.0875

Thus, mean number of apples in a box is 57.0875.

Question 2.

The daily expenditure of 50 hostel students are as follows:


Find the mean daily expenditure of the students of hostel using appropriate method.

Answer:

Let us solve by direct method since the frequencies have smaller values.

This is grouped frequency distribution.

We shall represent them by creating a table having columns of class intervals, midpoints, frequencies, and product of these midpoints and frequencies.

So now, we have

∑f_i = 50 and ∑x_if_i = 7260

For grouped frequency distribution, mean is taken out from the formula:

⇒  [from the table]

⇒ 

⇒ Mean = 145.2

Thus, the mean daily expenditure of the students is Rs. 145.2.

Question 3.

The mean of the following frequency distribution of 200 observations is 332. Find the value of x and y.

Answer:

This is a grouped frequency distribution.

To find x and y, we’ll need to find mean of the following distribution by assumed mean method and equate it to the given mean of the following distribution, 332.

And class size, c = 50.

[∵ (150 – 100) = (200 – 150) = (250 – 200) = … = (550 – 500) = 50]

This given data is in exclusive type.

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

So now, we have

∑f_iu_i = -130 – 3x

And ∑f_i = 146 + x + y = 200

Mean is given by

⇒  [given, mean = 332 and using the values from the table]

⇒ 

⇒ 

⇒ 43 × 4 = 130 + 3x

⇒ 172 = 130 + 3x

⇒ 3x = 172 – 130

⇒ 3x = 42

⇒ 

⇒ x = 14 …(i)

Since, there are 200 observations.

⇒ Sum of frequencies = 200 [Sum of frequencies depict total number of observation]

⇒ ∑f_i = 200

⇒ 146 + x + y = 200

⇒ x + y = 200 – 146

⇒ x + y = 54 …(ii)

Putting the value of x from equation (i) into equation (ii), we get

x + y = 54

⇒ 14 + y = 54

⇒ y = 54 – 14

⇒ y = 40

Thus, the missing frequencies are x = 14 and y = 40.

Question 4.

Find the mode of the following frequency distribution:

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 57 and the class corresponding to this frequency is 45 – 60.

So, this implies that,

Modal class = 45 – 60

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 45

f₀ = frequency of the class preceding the modal class = 23

f₁ = frequency of the modal class = 57

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

c = size of class interval (the class intervals are same) = 15

∴ Substituting the values l = 45, f₀ = 23, f₁ = 57, f₂ = 33 and c = 15 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ 

⇒ Mode = 45 + 8.793

⇒ Mode = 53.793

Thus, the mode of the data is 53.793.

Question 5.

Find the mode of the following data:

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 28 and the class corresponding to this frequency is 50 – 60.

So, this implies that,

Modal class = 50 – 60

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 50

f₀ = frequency of the class preceding the modal class = 17

f₁ = frequency of the modal class = 28

f₂ = frequency of the class succeeding the modal class = 23

c = size of class interval (the class intervals are same) = 10

∴ Substituting the values l = 50, f₀ = 17, f₁ = 28, f₂ = 23 and c = 10 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ Mode = 50 + 6.875

⇒ Mode = 56.875

Thus, the mode of the data is 56.875.

Question 6.

The mode of the following frequency distribution of 165 observations is 34.5. Find the value of a and b.

Answer:

Given that,

Total number of observations = 165

⇒ Sum of frequencies = 165

⇒ 5 + 11 + a + 53 + b + 16 + 10 = 165

⇒ 95 + a + b = 165

⇒ a + b = 165 – 95

⇒ a + b = 70 …(i)

Also, given that

Mode of the data = 34.5

Corresponding to the modal value, the class it lies between is 32 – 41. [∵ 34.5 lies between 32 – 41]

⇒ Modal class = 32 – 41

Now,

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 32

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

f₁ = frequency of the modal class = 53

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

c = size of class interval (the class intervals are same) = 9

∴ Substituting the values l = 32, f₀ = a, f₁ = 53, f₂ = b and c = 9 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ 2.5 × (106 – a – b) = 477 – 9a

⇒ 265 – 2.5a – 2.5b = 477 – 9a

⇒ 9a – 2.5a – 2.5b = 477 – 265

⇒ 6.5a – 2.5b = 212

⇒ 

⇒ 

⇒ 65a – 25b = 2120

⇒ 13a – 5b = 424 …(ii)

Multiply 5 by equation (i),

a + b = 70 [× 5

⇒ 5a + 5b = 350 …(iii)

Solving equation (i) and (iii), we get

13a – 5b = 424

5a + 5b = 350

18a + 0 = 774

⇒ 18a = 774

⇒ 

⇒ a = 43

Putting value a = 43 in equation (i),

a + b = 70

⇒ 43 + b = 70

⇒ b = 70 – 43

⇒ b = 27

Thus, a = 43 and b = 27.

Question 7.

Find the mode of the following frequency distribution:

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 86 and the class corresponding to this frequency is 3000 – 3500.

So, this implies that,

Modal class = 3000 – 3500

Mode of such grouped frequency distribution is given by,

Where,

l = lower limit of the modal class = 3000

f₀ = frequency of the class preceding the modal class = 60

f₁ = frequency of the modal class = 86

f₂ = frequency of the class succeeding the modal class = 74

c = size of class interval (the class intervals are same) = 500

∴ Substituting the values l = 3000, f₀ = 60, f₁ = 86, f₂ = 74 and c = 500 in the formula of mode. We get

⇒ 

⇒ 

⇒ 

⇒ 

⇒ Mode = 3000 + 342.10

⇒ Mode = 3342.10

Thus, the mode of the data is 3342.10.

Question 8.

Find the median of the following frequency distribution:

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 100

Now, we just need to find the value of n/2. So,

⇒ 

Now, look up for a value in the cumulative frequency just greater than 50.

We have, 59.

Corresponding to this value of cumulative frequency, we can say that median class is 40 – 50.

That is,

Median class = 40 – 50

∴ we have almost everything we require to calculate median.

Median is given by,

Where,

l = lower limit of the median class = 40

n = Total number of observation (sum of frequencies) = 100

cf = cumulative frequency of the class preceding the median class = 35

f = frequency of the median class = 24

c = class size (class sizes are equal) = 10

Putting the values, l = 40, n/2 = 50, cf = 35, f = 24 and c = 10 in the given formula of median, we get

⇒ 

⇒ 

⇒ Median = 40 + 6.25

⇒ Median = 46.25

Thus, the median of the data is 46.25.

Question 9.

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

Answer:

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 76 + x + y …(i)

But given is, sum of frequencies, ∑f = n = 100 …(ii)

⇒ 

⇒ 

Comparing equations (i) and (ii), we get

76 + x + y = 100

⇒ x + y = 100 – 76

⇒ x + y = 24 …(iii)

Also, given that, median = 525

Corresponding to this value of median, we can say that median lies in the class is 500 – 600. [∵ 525 lies between 500 – 600]

⇒ Median class = 500 – 600

Median is given by,

Where,

l = lower limit of the median class = 500

n = Total number of observation (sum of frequencies) = 100

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

f = frequency of the median class = 20

c = class size (class sizes are equal) = 100

Putting the values, l = 500, n/2 = 50, cf = 36 + x, f = 20 and c = 100 in the given formula of median, we get

⇒  [∵ given that, median = 525]

⇒ 

⇒ 

⇒ (14 – x) × 5 = 25

⇒ 14 – x = 5

⇒ x = 14 – 5

⇒ x = 9

Substituting the value, x = 9 in equation (iii), we get

x + y = 24

⇒ 9 + y = 24

⇒ y = 24 – 9

⇒ y = 15

Thus, x = 9 and y = 15.

Question 10.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

For some data, if Z = 25 and  = 25, then M =

A. 25

B. 75

C. 50

D. 0

Answer:

Given: Z = 25 [Z denotes mode]

 [ denotes mean]

We have to find M (that is, median).

By empirical relationship, we can say

Mode = 3 (Median) – 2 (Mean)

⇒ 

⇒ 

⇒ 

Putting in the given values, we can

⇒ 

⇒ 

⇒ M = 25

Thus, option (a) is correct.

Question 11.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

For some data Z — M = 2.5. If the mean of the data is 20, then Z =

A. 21.25

B. 22.75

C. 23.75

D. 22.25

Answer:

We have been given that,

Z – M = 2.5

⇒ M = Z – 2.5 …(i)

And  …(ii)

Here, Z = Mode,

M = Median

& 

So, by empirical relationship we have,

Mode = 3 (Median) – 2 (Mean)

⇒ 

Putting equations (i) and (ii) in the above equation, we get

Z = 3(Z – 2.5) – 2×20

⇒ Z = 3Z – 7.5 – 40

⇒ 3Z – Z = 40 + 7.5

⇒ 2Z = 47.5

⇒ 

⇒ Z = 23.75

Thus, option (c) is correct.

Question 12.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If — Z = 3 and  + Z = 45, then M =

A. 24

B. 22

C. 26

D. 23

Answer:

Solving the given equations,

⇒ 

⇒ 

⇒ 

Put this value in ,

24 – Z = 3

⇒ Z = 24 – 3

⇒ Z = 21

By empirical relationship, we can write

Mode = 3 (Median) – 2 (Mean)

⇒ 

⇒ 

⇒ 

Putting in the given values, we can

⇒ 

⇒ 

⇒ M = 23

Thus, option (d) is correct.

Question 13.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If Z = 24,  = 18, then M=

A. 10

B. 20

C. 30

D. 40

Answer:

Given that,

Z = 24

To find: M = ?

By empirical relationship, we can write

Mode = 3 (Median) – 2 (Mean)

⇒ 

⇒ 

⇒ 

Putting in the given values, we can

⇒ 

⇒ 

⇒ M = 20

Thus, option (b) is correct.

Question 14.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If M = 15, = 10, then Z=

A. 15

B. 20

C. 25

D. 30

Answer:

Given that,

M = 15

So, by empirical relationship we have,

Mode = 3 (Median) – 2 (Mean)

⇒ 

Putting given values in the above equation, we get

Z = 3(15) – 2×10

⇒ Z = 45 – 20

⇒ Z = 25

Thus, option (c) is correct.

Question 15.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

(6) If M = 22, Z = 16, then  =

A. 22

B. 25

C. 32

D. 66

Answer:

Given that,

M = 22

Z = 16

So, by empirical relationship we have,

Mode = 3 (Median) – 2 (Mean)

⇒ 

⇒ 

⇒ 

Putting in the given values in the above equation, we get

⇒ 

⇒ 

⇒ 

Thus, option (b) is correct.

Question 16.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

(7) If = 21.44 and Z = 19.13, then M =

A. 21.10

B. 19.67

C. 20.10

D. 20.67

Answer:

Given that,

Z = 19.13

To find: M = ?

By empirical relationship, we can write

Mode = 3 (Median) – 2 (Mean)

⇒ 

⇒ 

⇒ 

Putting in the given values, we can

⇒ 

⇒ 

⇒ M = 20.67

Thus, option (d) is correct.

Question 17.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If M = 26,  =36, then Z =

A. 6

B. 5

C. 4

D. 3

Answer:

Given that,

M = 15

So, by empirical relationship we have,

Mode = 3 (Median) – 2 (Mean)

⇒ 

Putting given values in the above equation, we get

Z = 3(15) – 2×10

⇒ Z = 45 – 20

⇒ Z = 25

Thus, option (c) is correct.

Question 18.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

(9) The modal class of the frequency distribution given below is


A. 10-20

B. 20-30

C. 30-40

D. 40-50

Answer:

Observe that, from the given data:

The maximum class frequency, here, is 17 and the class corresponding to this frequency is 30 – 40.

So, this implies that,

Modal class = 30 – 40

[∵ modal class in a grouped frequency table can be calculated by corresponding maximum class frequency]

Thus, option (c) is correct.

Question 19.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The cumulative frequency of class 20-30 of the frequency distribution given in (9) is ….

A. 25

B. 35

C. 15

D. 40

Answer:

We have

Cumulative frequency of a class is just the sum of frequencies of the particular class along with all the classes preceding the class.

So, Cumulative frequency of class 20 – 30 = Sum of frequency of the class 20 – 30 along with frequencies of all the classes preceding the class 20 – 30.

⇒ Cumulative frequency of class 20 – 30 = 7 + 15 + 13

⇒ Cumulative frequency of class 20 – 30 = 35

Thus, option (b) is correct.

Question 20.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The median class of the frequency distribution given in (9) is ….

A. 40-50

B. 30-40

C. 20-30

D. 10-20

Answer:

We have

We need to show it into a proper table representing frequencies and cumulative frequencies.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 62

Now, we just need to find the value of n/2. So,

⇒ 

Now, look up for a value in the cumulative frequency just greater than 31.

We have, 35.

Corresponding to this value of cumulative frequency, we can say that median class is 20 – 30.

That is,

Median class = 20 – 30

Thus, option (c) is correct.