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 
5052  6 
5254  14 
5456  16 
5658  9 
5860  5 
Solution:
Number of Mangoes  Midpoint \( (x_i) \)  Number of Boxes \( (f_i) \)  \( f_i x_i \) 
5052  51  6  306 
5254  53  14  742 
5456  55  16  880 
5658  57  9  513 
5860  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)  1020  2030  3040  4050  5060  6070 
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 \((f_i)\)  \( f_i x_i \) 
1020  15  12  180 
2030  25  8  200 
3040  35  22  770 
4050  45  20  900 
5060  55  18  990 
6070  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  010  1020  2030  3040  4050 
Frequency  4  6  10  6  4 
Solution:
Class Interval  Midpoint \((x_i)\)  Frequency \((f_i)\)  \( f_i x_i \) 
010  5  4  20 
1020  15  6  90 
2030  25  10  250 
3040  35  6  210 
4050  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  1020  2030  3040  4050  5060  6070 
Frequency  10  16  20  30  13  11 
Solution:
Class Interval  Midpoint \((x_i)\)  Frequency \((f_i)\)  \( f_i x_i \) 
1020  15  10  150 
2030  25  16  400 
3040  35  20  700 
4050  45  30  1350 
5060  55  13  715 
6070  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  040  4080  80120  120160  160200 
Frequency  12  20  25  20  13 
Solution:
Class Interval  Midpoint \((x_i)\)  Frequency \((f_i)\)  Assumed Mean (A)  Deviation \((d_i=x_i – A) \)  \( f_i d_i \) 
040  20  12  100  80  960 
4080  60  20  100  40  800 
80120  100  25  100  0  0 
120160  140  20  100  40  800 
160200  180  13  100  80  1040 
Total  90  80 
Here’s how we calculate the mean using the assumed mean method:

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

Calculate Deviations (d): Subtract the assumed mean (A) from each midpoint \((x_i)\): \( d_i=x_iA \)
 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  2535  3545  4555  5565  6575 
Frequency  4  10  8  12  6 
Solution:
Class Interval  Midpoint \((x_i)\)  Frequency \((f_i)\)  Assumed Mean (A)  Deviation \((d_i=x_i – A) \)  \( f_i d_i \) 
2535  30  4  50  20  80 
3545  40  10  50  10  100 
4555  50  8  50  0  0 
5565  60  12  50  10  120 
6575  70  6  50  20  120 
Total  40  60 
Here’s the breakdown:

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

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

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  030  3060  6090  90120  120150 
Frequency  12  15  20  25  8 
Solution:
Class Interval  Midpoint \((x_i)\)  Frequency \((f_i)\)  Assumed Mean (A)  Step Deviation \( (u_i =\frac{x_i – A}{h} )\)  \( f_i u_i \) 
030  15  12  60  3  36 
3060  45  15  60  1  15 
6090  75  20  60  1  20 
90120  105  25  60  3  75 
120150  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  014  1428  2842  4256  5670 
Frequency  7  21  35  11  16 
Solution:
Class Interval  Midpoint \((x_i)\)  Frequency \((f_i)\)  Assumed Mean (A)  Step Deviation \( (u_i =\frac{x_i – A}{h}) \)  \( f_i u_i \) 
014  7  7  28  3  21 
1428  21  21  28  1  21 
2842  35  35  28  1  35 
4256  49  11  28  3  33 
5670  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)  010  1020  2030  3040  4050 
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 \((f_i)\)  \(f_i x_i \) 
010  5  15  75 
1020  15  20  300 
2030  25  35  875 
3040  35  p  35×p 
4050  45  10  450 
80+p  1700+35p 
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:
Age (years)  3034  3539  4044  4549  5054  5559 
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 \((f_i)\)  \( f_i x_i \) 
3034  29.5 – 34.5  32  10  320 
3539  34.5 – 39.5  37  12  444 
4044  39.5 – 44.5  42  15  630 
4549  44.5 – 49.5  47  6  282 
5054  49.5 – 54.5  52  4  208 
5559  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  514  1524  2534  3544  4554  5564 
Frequency  3  6  18  20  10  3 
Solution:
Create Exclusive Class Limits and Midpoints:
Class Interval  Exclusive Class Limits  Midpoint \((x_i)\)  Frequency \((f_i)\)  \( f_i x_i \) 
514  4.5 – 14.5  9.5  3  28.5 
1524  14.5 – 24.5  19.5  6  117 
2534  24.5 – 34.5  29.5  18  531 
3544  34.5 – 44.5  39.5  20  790 
4554  44.5 – 54.5  49.5  10  495 
5564  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 \((x_i) \)  \( f_i x_i \) 
010  5  5  25 
1020  4 (=9 − 5)  15  60 
2030  8 (=17 − 9)  25  200 
3040  12 (=29 − 17)  35  420 
4050  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:

Solution:
Class Interval (marks)  Midpoint \((x_i) \)  Students \((f_i) \)  \( f_i x_i \) 
14  2.5  6  15 
49  6.5  12  78 
916  12.5  26  325 
1617  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.