WBCHSE – Let us Work Out 26.1 – WB Class 10

Q1) I have written ages of my 40 friends in the table given below.

Let us find the average age of my friends by the direct method.

Solution: 

Calculate the Average:

Formula: Average \( (x̄) = \frac{\sum x_if_i}{\sum f_i}\)

Substitute: \( x̄ = \frac{697}{40} = 17.425 = 17.43\) (approx.)

Therefore, the average age of my friends is 17.43 years.

Q2) I have written member of 50 families of our village in the table given below.

Let us write the average member of 50 families by the method of assumed mean.

Solution: 

Calculate the Average:

• Formula: Average \( (x̄) = A +  \frac{\sum f_i \times d_i}{\sum f_i}\)

• Substitute: \( x̄ = 4 + \frac{12}{50} = 4 + 0.24 = 4.24\)

Therefore, the average number of members per family in the village is 4.24.

Q3) If the arithmetic mean of the data given below is 20.6, let us find the value of ‘a’.

Solution:

Use the Table and Formula:

• We know the average \( x̄ = 20.6\) and the total frequency \(\sum f_i  = 50\).

• We also see from the table that \( \sum x_i f_i = 530 + 25a\).

Now we can plug these values directly into the arithmetic mean formula, we get

\( x̄ = \frac{\sum x_i f_i}{\sum f_i}\)

\(\Rightarrow 20.6 = \frac{530 + 25a}{50}\)

\(\Rightarrow 20.6 × 50 = 530 + 25a \)

\(\Rightarrow 1030 = 530 + 25a \)

\(\Rightarrow 1030 – 530 = 25a \)

\(\Rightarrow 500 = 25a \)

\(\Rightarrow a = \frac{500}{50} \)

\(\therefore a = 20 \)

Therefore, the value of ‘a’ is 20.

Q4) If the arithmetic mean of the distribution given below is 15, let us find the value of p.

Solution:

Use the Table and Formula:

• We know the average \(x̄ = 15 \) and the total frequency \( \sum f_i = 27 + p\).

• We also see from the table that \( \sum x_i f_i = 445 + 10p \).

Now we can plug these values directly into the arithmetic mean formula:

\( x̄ = \frac{\sum x_i f_i}{\sum f_i}\)

∴ \( 15 = \frac{445 + 10p}{27 + p}\)

Solve for ‘p’:

⇒ 15(27 + p) = 445 + 10p

⇒ 405 + 15p = 445 + 10p

⇒ 15p – 10p = 445 – 405

⇒ 5p = 40

⇒ p = 40 / 5

∴ p = 8

Therefore, the value of ‘p’ is 8.

Q5) Rahamatchacha will go to the retail market to sell mangoes kept in 50 packing boxes. The table below shows the distribution of boxes based on the number of mangoes they contain. Find the mean number of mangoes per box.

Solution:

Mean \(\overline{x}= \frac{\sum f_i x_i}{\sum f_i } = \frac{2736}{50} = 54.72 \)

Therefore, the mean number of mangoes per box is 54.72.

Q6) Mahidul has recorded the ages of 100 patients at a village hospital. Calculate the average age of these 100 patients using the data provided below:

Solution:

We’ll use the following table to calculate the average age:

Average Age = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{4340}{100} = 43.4 \ \text{years}\)

Therefore, the average age of the 100 patients at the village hospital is 43.4 years.

Q7) (i) Let us find the mean of the following data by the direct method:

Solution:

Mean \( \overline{x} = \frac{\sum f_i x_i}{\sum f_i}  = \frac{750}{30} = 25\)

Therefore, the mean of the given data is 25.

Q7 (ii) Let us find the mean of the following data by the direct method: 

Solution:

Mean \(\overline{x} = \frac{\sum f_i x_i}{\sum f_i} \) = 4030 / 100 = 40.3

Therefore, the mean of the given data is 40.3.

Q8 (i) Let us find the mean of the following data by the assumed mean method:

Solution:

Here’s how we calculate the mean using the assumed mean method:

1. Choose an Assumed Mean (A): We choose a convenient midpoint from the data, in this case, 100.

2. Calculate Deviations (d): Subtract the assumed mean (A) from each midpoint \((x_i)\): \( d_i=x_i-A \)

3. Calculate the Mean:
   Mean \( \overline{x}=A+ \frac{\sum f_i d_i}{\sum f_i} \)
   ⇒ \( \overline{x} \) = 100 + (80 / 90)
   ⇒ \( \overline{x} \) = 100 + 0.89
   Mean \( \overline{x} \) = 100.89

Therefore, the mean of the given data, calculated using the assumed mean method, is 100.89.

Q8 (ii) Let us find the mean of the following data by the assumed mean method:

Solution: 

Here’s the breakdown:

1. Choose an Assumed Mean (A): We choose a convenient midpoint from the data. In this case, we’ll use 50.

2. Calculate Deviations (d): Subtract the assumed mean (A) from each midpoint \((x_i)\): \( d_i=x_i-A \)

3. Calculate the Mean:
   Mean \( \overline{x}=A+ \frac{\sum f_i d_i}{\sum f_i} \)
   Mean \( \overline{x} \) = 50 + (60 / 40)
   Mean \( \overline{x} \) = 50 + 1.5
   Mean \( \overline{x} \) = 51.5

Therefore, using the assumed mean method, the mean of the data is confirmed to be 51.5.

Q9 (i) Find the mean of the following data using the step deviation method:

Solution:

Where:

• h (class width) = 30

Mean \( \overline{x} = A + h \times (\frac{\sum f_i u_i}{\sum f_i})\)
Mean \( \overline{x} \) = 60 + 30 × (84 / 80)
Mean \( \overline{x} \) = 91.5

Therefore, the mean is 91.5.

Q9 (ii) Find the mean of the following data using the step deviation method:

Solution:

Where:

• h (class width) = 14

Mean \( \overline{x} = A + h \times (\frac{\sum f_i u_i}{\sum f_i})\)
Mean \( \overline{x} \) = 28 + 14 × (106/ 90)
Mean \( \overline{x} \) = 44.49

Therefore, the mean is 44.49.

Q10) If the mean of the following frequency distribution table is 24, find the value of p.

Solution:

We can use the formula for calculating the mean of grouped data:

Mean \( \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \)

where:

• \(\overline{x} \) is the mean

• \( f \) is the frequency of each class

• \( x \) is the midpoint of each class

1. Calculate the midpoints \((x_i)\) of each class interval:

2. Simplify and solve for p:

\( 24 = \frac{1700 + 35p}{80 + p} \)
⇒ 24(80 + p) = 1700 + 35p
⇒ 1920 + 24p = 1700 + 35p
⇒ 220 = 11p
∴ p = 20

Therefore, the value of p is 20.

Q11) Let us see the ages of the persons present in a meeting and determine their average age from the following table:

Solution:

1. Create Exclusive Class Limits and Midpoints:

2. Calculate the Mean:

Average Age ​\( \overline{x}= \frac{\sum f_i x_i} {\sum f_i} = \frac{2055}{50} = 41.1 \text{years} \)​

Therefore, even with the adjustment to exclusive class limits, the average age of the persons in the meeting remains 41.1 years.

Key Point: While converting to exclusive class limits is crucial for accuracy, especially for finding the mode, in this specific case, it did not affect the calculated mean. This is because the adjustments to the class limits resulted in the same midpoint values.

Answer: The average age of the persons present in the meeting is 41.1 years.

Q12) Let us find the mean of the following data:

Solution:

Create Exclusive Class Limits and Midpoints:

Mean \(\overline{x}= \frac{\sum f_i x_i} {\sum f_i} = \frac{2140}{60} = 35.67 \)

Answer: The mean of the given data is 35.67.

Q13) Let us find the mean of obtaining marks of girl students if their cumulative frequencies are as follows:

Solution:
First, we need to convert the cumulative frequencies to regular frequencies:

Mean \(\overline{x}= \frac{\sum f_i x_i} {\sum f_i} = \frac{1425}{45} = 31.67\)

Answer: The mean of the marks obtained by the girl students is 31.67.

Q14) Let us find the mean of the obtaining marks of 64 students from the table given below:

Solution:

Mean \(\overline{x}= \frac{\sum f_i x_i } {\sum f_i} = \frac{748}{64} = 11.69 \)

Answer: The mean of the marks obtained by the 64 students is 11.69.