##### Class 10^{th} Mathematics Gujarat Board Solution

**Exercise 15.1**- Find the mean of the following frequency
- Find the mean wage of 200 workers of a factory where wages are classified as…
- Marks obtained by 140 students of class X out of 50 in mathematics are given in…
- Find the mean of the following frequency distribution by step-deviation…
- Find the mean for the following frequency
- A survey conducted by a student of B.B.A. for daily income of 600 families is…
- The number of shares held by a person of various companies are as follows. Find…
- The mean of the following frequency distribution of 100 observations is 148.…
- The table below gives the percentage of girls in higher secondary science…
- The following distribution shows the number of out door patients in 64…

**Exercise 15.2**- Find the mode for the following frequency
- The data obtained for 100 shops for their daily profit per shop are as…
- Daily wages of 90 employees of a factory are as follows:Daily wages (in…
- Find the mode for the following
- Find the mode for the following
- The following data gives the information of life of 200 electric bulbs (in…

**Exercise 15.3**- Find the median for the following:Value of
- Find the median for the following frequency
- Find the median from following frequency
- The following frequency distribution represents the deposits (in thousand…
- The median of the following frequency distribution is 38. Find the value of a…
- The median of 230 observations of the following frequency distribution is 46.…
- The following table gives the frequency distribution of marks scored by 50…

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

**Exercise 15.1**

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

**Exercise 15.2**

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

**Exercise 15.3**

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

**Exercise 15**

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

**Answer:**This is a grouped frequency distribution.

To find f_{1} and f_{2}, 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_{i}f_{i} = 8869 + 124.5f_{1} + 274.5f_{2}

And ∑f_{i} = 62 + f_{1} + f_{2}

Mean is given by

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

⇒ 148 × (62 + f_{1} + f_{2}) = 8869 + 124.5f_{1} + 274.5f_{2}

⇒ 9176 + 148f_{1} + 148f_{2} = 8869 + 124.5f_{1} + 274.5f_{2}

⇒ 274.5f_{2} – 148f_{2} + 124.5f_{1} – 148f_{1} = 9176 – 8869

⇒ 126.5f_{2} – 23.5f_{1} = 307

⇒

⇒

⇒ 253f_{2} – 47f_{1} = 307 × 2

⇒ 253f_{2} – 47f_{1} = 614 …(i)

Since, there are 100 observations.

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

⇒ ∑f_{i} = 100

⇒ 62 + f_{1} + f_{2} = 100

⇒ f_{1} + f_{2} = 100 – 62 = 38

⇒ f_{1} + f_{2} = 38 …(ii)

Multiplying 47 by equation (ii), we get

f_{1} + f_{2} = 38 [× 47

⇒ 47f_{1} + 47f_{2} = 1786 …(iii)

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

253f_{2} – 47f_{1} = 614

47f_{2} + 47f_{1} = 1786

300f_{2} + 0 = 2400

⇒ 300f_{2} = 2400

⇒

⇒ f_{2} = 8

Putting f_{2} = 8 in equation (ii), we get

f_{1} + 8 = 38

⇒ f_{1} = 38 – 8

⇒ f_{1} = 30

**Thus, the missing frequencies are f**_{1} = 30 and f_{2} = 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_{i}u_{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 f_{1} and f_{2} :

**Answer:**This is a grouped frequency distribution.

To find f_{1} and f_{2}, 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_{i}f_{i} = 608 + 16f_{1} + 20f_{2}

And ∑f_{i} = 35 + f_{1} + f_{2}

Mean is given by

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

⇒ 18 × (35 + f_{1} + f_{2}) = 608 + 16f_{1} + 20f_{2}

⇒ 630 + 18f_{1} + 18f_{2} = 608 + 16f_{1} + 20f_{2}

⇒ 20f_{2} – 18f_{2} + 16f_{1} – 18f_{1} = 630 – 608

⇒ 2f_{2} – 2f_{1} = 22

⇒ 2(f_{2} – f_{1}) = 22

⇒ f_{2} – f_{1} = 11 …(i)

Since, there are 64 hospitals.

⇒ Number of hospitals = 64

⇒ Frequency = 64

⇒ ∑f_{i} = 64

⇒ 35 + f_{1} + f_{2} = 64

⇒ f_{1} + f_{2} = 64 – 35

⇒ f_{1} + f_{2} = 29 …(ii)

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

f_{2} – f_{1} = 11

f_{2} + f_{1} = 29

2f_{2} + 0 = 40

⇒ 2f_{2} = 40

⇒

⇒ f_{2} = 20

Putting f_{2} = 20 in equation (ii), we get

f_{1} + 20 = 29

⇒ f_{1} = 29 – 20

⇒ f_{1} = 9

**Thus, the missing frequencies are f**_{1} = 9 and f_{2} = 20.

**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_{i}f_{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_{i}f_{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_{i}d_{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_{i}u_{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_{i}f_{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_{i}d_{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_{i}f_{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 f_{1} and f_{2} :

**Answer:**

This is a grouped frequency distribution.

To find f_{1} and f_{2}, 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_{i}f_{i} = 8869 + 124.5f_{1} + 274.5f_{2}

And ∑f_{i} = 62 + f_{1} + f_{2}

Mean is given by

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

⇒ 148 × (62 + f_{1} + f_{2}) = 8869 + 124.5f_{1} + 274.5f_{2}

⇒ 9176 + 148f_{1} + 148f_{2} = 8869 + 124.5f_{1} + 274.5f_{2}

⇒ 274.5f_{2} – 148f_{2} + 124.5f_{1} – 148f_{1} = 9176 – 8869

⇒ 126.5f_{2} – 23.5f_{1} = 307

⇒

⇒

⇒ 253f_{2} – 47f_{1} = 307 × 2

⇒ 253f_{2} – 47f_{1} = 614 …(i)

Since, there are 100 observations.

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

⇒ ∑f_{i} = 100

⇒ 62 + f_{1} + f_{2} = 100

⇒ f_{1} + f_{2} = 100 – 62 = 38

⇒ f_{1} + f_{2} = 38 …(ii)

Multiplying 47 by equation (ii), we get

f_{1} + f_{2} = 38 [× 47

⇒ 47f_{1} + 47f_{2} = 1786 …(iii)

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

253f_{2} – 47f_{1} = 614

47f_{2} + 47f_{1} = 1786

300f_{2} + 0 = 2400

⇒ 300f_{2} = 2400

⇒

⇒ f_{2} = 8

Putting f_{2} = 8 in equation (ii), we get

f_{1} + 8 = 38

⇒ f_{1} = 38 – 8

⇒ f_{1} = 30

**Thus, the missing frequencies are f _{1} = 30 and f_{2} = 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_{i}u_{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 f_{1} and f_{2} :

**Answer:**

This is a grouped frequency distribution.

To find f_{1} and f_{2}, 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_{i}f_{i} = 608 + 16f_{1} + 20f_{2}

And ∑f_{i} = 35 + f_{1} + f_{2}

Mean is given by

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

⇒ 18 × (35 + f_{1} + f_{2}) = 608 + 16f_{1} + 20f_{2}

⇒ 630 + 18f_{1} + 18f_{2} = 608 + 16f_{1} + 20f_{2}

⇒ 20f_{2} – 18f_{2} + 16f_{1} – 18f_{1} = 630 – 608

⇒ 2f_{2} – 2f_{1} = 22

⇒ 2(f_{2} – f_{1}) = 22

⇒ f_{2} – f_{1} = 11 …(i)

Since, there are 64 hospitals.

⇒ Number of hospitals = 64

⇒ Frequency = 64

⇒ ∑f_{i} = 64

⇒ 35 + f_{1} + f_{2} = 64

⇒ f_{1} + f_{2} = 64 – 35

⇒ f_{1} + f_{2} = 29 …(ii)

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

f_{2} – f_{1} = 11

f_{2} + f_{1} = 29

2f_{2} + 0 = 40

⇒ 2f_{2} = 40

⇒

⇒ f_{2} = 20

Putting f_{2} = 20 in equation (ii), we get

f_{1} + 20 = 29

⇒ f_{1} = 29 – 20

⇒ f_{1} = 9

**Thus, the missing frequencies are f _{1} = 9 and f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 7

f_{1} = frequency of the modal class = 15

f_{2} = 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, f_{0} = 7, f_{1} = 15, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 18

f_{1} = frequency of the modal class = 27

f_{2} = 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, f_{0} = 18, f_{1} = 27, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 12

f_{1} = frequency of the modal class = 33

f_{2} = 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, f_{0} = 12, f_{1} = 33, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 42

f_{1} = frequency of the modal class = 82

f_{2} = 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, f_{0} = 42, f_{1} = 82, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 21

f_{1} = frequency of the modal class = 31

f_{2} = 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, f_{0} = 21, f_{1} = 31, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 42

f_{1} = frequency of the modal class = 82

f_{2} = 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, f_{0} = 42, f_{1} = 82, f_{2} = 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.**

**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

f_{0} = frequency of the class preceding the modal class = 7

f_{1} = frequency of the modal class = 15

f_{2} = 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, f_{0} = 7, f_{1} = 15, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 18

f_{1} = frequency of the modal class = 27

f_{2} = 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, f_{0} = 18, f_{1} = 27, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 12

f_{1} = frequency of the modal class = 33

f_{2} = 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, f_{0} = 12, f_{1} = 33, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 42

f_{1} = frequency of the modal class = 82

f_{2} = 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, f_{0} = 42, f_{1} = 82, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 21

f_{1} = frequency of the modal class = 31

f_{2} = 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, f_{0} = 21, f_{1} = 31, f_{2} = 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

f_{0} = frequency of the class preceding the modal class = 42

f_{1} = frequency of the modal class = 82

f_{2} = 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, f_{0} = 42, f_{1} = 82, f_{2} = 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,

**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,