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Jul 23, 2024

# WBCHSE – Let us Work Out 26.1 – WB Class 10

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

 Age (Years) 15 16 17 18 19 20 Number of Friends 4 7 10 10 5 4

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

Solution:

 Age (Years) $$(x_i)$$ Number of friends (frequency) $$(f_i)$$ $$x_if_i$$ 15 4 15 $$\times$$ 4 = 60 16 7 16 $$\times$$ 7 = 112 17 10 17 $$\times$$ 10 = 170 18 10 18 $$\times$$ 10 = 180 19 5 19 $$\times$$ 5 = 95 20 4 20 $$\times$$ 4 = 80 Total $$\sum f_i = 40$$ $$\sum x_if_i = 697$$

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.

 Number of members 2 3 4 5 6 7 Number of families 6 8 14 15 4 3

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

Solution:

 Number of Members $$(x_i)$$ Number of Families $$(f_i)$$ Deviation $$(d_i = x_i − A)$$ $$f_i \cdot d_i$$ 2 6 2 − 4 = −2 6 $$\times$$ (−2\) = −12 3 8 3 − 4 = −1 8 $$\times$$ (−1\) = −8 4 14 4 − 4 = 0 14 $$\times$$ 0 = 0 5 15 5 − 4 = 1 15 $$\times$$ 1 = 15 6 4 6 − 4 = 2 4 $$\times$$ 2 = 8 7 3 7 − 4 = 3 3 $$\times$$ 3 = 9 Total $$\sum f_i = 50$$ $$\sum f_i \cdot d_i = 12$$

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’.

 Variables $$(x_i)$$ 10 15 a 25 35 Frequency $$(f_i)$$ 3 10 25 7 5

Solution:

 Variables $$(x_i)$$ Frequency $$(f_i)$$ $$x_i f_i$$ 10 3 10 × 3 = 30 15 10 15 × 10 = 150 a 25 a × 25 = 25a 25 7 25 × 7 = 175 35 5 35 × 5 = 175 Total $$\sum f_i = 50$$ $$\sum x_i f_i = 530 + 25a$$

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.

 Score $$(x_i)$$ 5 10 15 20 25 Frequency $$(f_i)$$ 6 p 6 10 5

Solution:

 Variable $$(x_i)$$ Frequency $$(f_i)$$ $$x_i f_i$$ 5 6 5 × 6 = 30 10 p 10 × p = 10p 15 6 15 × 6 = 90 20 10 20 × 10 = 200 25 5 25 × 5 = 125 Total $$\sum f_i = 27 + p$$ $$\sum x_i f_i = 445 + 10p$$

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.

 Number of Mangoes Number of Boxes 50-52 6 52-54 14 54-56 16 56-58 9 58-60 5

Solution:

 Number of Mangoes Midpoint $$(x_i)$$ Number of Boxes $$(f_i)$$ $$f_i x_i$$ 50-52 51 6 306 52-54 53 14 742 54-56 55 16 880 56-58 57 9 513 58-60 59 5 295 Total $$\sum f_i = 50$$ $$\sum f_i =2736$$

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:

 Age (years) 10-20 20-30 30-40 40-50 50-60 60-70 Number of patients 12 8 22 20 18 20

Solution:

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

 Age (years) Midpoint $$(x_i)$$ Number of patients $$(f_i)$$ $$f_i x_i$$ 10-20 15 12 180 20-30 25 8 200 30-40 35 22 770 40-50 45 20 900 50-60 55 18 990 60-70 65 20 1300 Total 100 4340

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:

 Class Interval 0-10 10-20 20-30 30-40 40-50 Frequency 4 6 10 6 4

Solution:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ $$f_i x_i$$ 0-10 5 4 20 10-20 15 6 90 20-30 25 10 250 30-40 35 6 210 40-50 45 4 180 Total 30 750

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:

 Class interval 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 10 16 20 30 13 11

Solution:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ $$f_i x_i$$ 10-20 15 10 150 20-30 25 16 400 30-40 35 20 700 40-50 45 30 1350 50-60 55 13 715 60-70 65 11 715 Total 100 4030

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:

 Class Interval 0-40 40-80 80-120 120-160 160-200 Frequency 12 20 25 20 13

Solution:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ Assumed Mean (A) Deviation $$(d_i=x_i – A)$$ $$f_i d_i$$ 0-40 20 12 100 -80 -960 40-80 60 20 100 -40 -800 80-120 100 25 100 0 0 120-160 140 20 100 40 800 160-200 180 13 100 80 1040 Total 90 80

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:

 Class Interval 25-35 35-45 45-55 55-65 65-75 Frequency 4 10 8 12 6

Solution:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ Assumed Mean (A) Deviation $$(d_i=x_i – A)$$ $$f_i d_i$$ 25-35 30 4 50 -20 -80 35-45 40 10 50 -10 -100 45-55 50 8 50 0 0 55-65 60 12 50 10 120 65-75 70 6 50 20 120 Total 40 60

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:

 Class Interval 0-30 30-60 60-90 90-120 120-150 Frequency 12 15 20 25 8

Solution:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ Assumed Mean (A) Step Deviation $$(u_i =\frac{x_i – A}{h} )$$ $$f_i u_i$$ 0-30 15 12 60 -3 -36 30-60 45 15 60 -1 -15 60-90 75 20 60 1 20 90-120 105 25 60 3 75 120-150 135 8 60 5 40 Total 80 84

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:

 Class Interval 0-14 14-28 28-42 42-56 56-70 Frequency 7 21 35 11 16

Solution:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ Assumed Mean (A) Step Deviation $$(u_i =\frac{x_i – A}{h})$$ $$f_i u_i$$ 0-14 7 7 28 -3 -21 14-28 21 21 28 -1 -21 28-42 35 35 28 1 35 42-56 49 11 28 3 33 56-70 63 16 28 5 80 Total 90 106

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.

 Class Interval (Number) 0-10 10-20 20-30 30-40 40-50 Number of students 15 20 35 p 10

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:

 Class Interval Midpoint $$(x_i)$$ Frequency $$(f_i)$$ $$f_i x_i$$ 0-10 5 15 75 10-20 15 20 300 20-30 25 35 875 30-40 35 p 35×p 40-50 45 10 450 80+p 1700+35p

2. Simplify and solve for p:

$$24 = \frac{1700 + 35p}{80 + p}$$
⇒ 24(80 + p) = 1700 + 35
p

⇒ 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:

 Age (years) 30-34 35-39 40-44 45-49 50-54 55-59 Number of persons 10 12 15 6 4 3

Solution:

1. Create Exclusive Class Limits and Midpoints:

 Age (years) Exclusive Class Limits Midpoint $$(x_i)$$ Number of Persons $$(f_i)$$ $$f_i x_i$$ 30-34 29.5 – 34.5 32 10 320 35-39 34.5 – 39.5 37 12 444 40-44 39.5 – 44.5 42 15 630 45-49 44.5 – 49.5 47 6 282 50-54 49.5 – 54.5 52 4 208 55-59 54.5 – 59.5 57 3 171 Total 50 2055

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:

 Class Interval 5-14 15-24 25-34 35-44 45-54 55-64 Frequency 3 6 18 20 10 3

Solution:

Create Exclusive Class Limits and Midpoints:

 Class Interval Exclusive Class Limits Midpoint $$(x_i)$$ Frequency $$(f_i)$$ $$f_i x_i$$ 5-14 4.5 – 14.5 9.5 3 28.5 15-24 14.5 – 24.5 19.5 6 117 25-34 24.5 – 34.5 29.5 18 531 35-44 34.5 – 44.5 39.5 20 790 45-54 44.5 – 54.5 49.5 10 495 55-64 54.5 – 64.5 59.5 3 178.5 Total 60 2140

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:

 Class Interval (marks) Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Number of girl students 5 9 17 29 45

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

 Class Interval (marks) Frequency $$(f_i)$$ Midpoint $$(x_i)$$ $$f_i x_i$$ 0-10 5 5 25 10-20 4 (=9 − 5) 15 60 20-30 8 (=17 − 9) 25 200 30-40 12 (=29 − 17) 35 420 40-50 16 (=45 − 29) 45 720 Total 45 1425

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:

 Class Interval (marks) 1-4 4-9 9-16 16-17 Students 6 12 26 20

Solution:

 Class Interval (marks) Midpoint $$(x_i)$$ Students $$(f_i)$$ $$f_i x_i$$ 1-4 2.5 6 15 4-9 6.5 12 78 9-16 12.5 26 325 16-17 16.5 20 330 Total 64 748

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.

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