Q1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

Solution: We use the direct method because the values of frequency (fi) and class marks (xi) are small.

Class mark is calculated using:

\[ x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2} \]

Mean: \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1 \]

Therefore, the mean number of plants per house is 8.1.

Q2. Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution: Use assumed mean (step-deviation) method. Choose a = 550, h = 20.

Mean formula: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \], where \( u_i = \frac{x_i – a}{h} \)

Mean: \[ \bar{x} = 550 + 20 \cdot \frac{-12}{50} = 550 – 4.8 = 545.20 \]

Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Solution: Class marks xi = (Upper + Lower)/2. Mean formula: \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]

Solving: 18 = (752 + 20f)/(44 + f) → f = 20

Q4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded. Find the mean heartbeats per minute.

Solution: Use assumed mean method. a = 75.5, h = 3, Mean: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \]

Mean: \[ \bar{x} = 75.5 + 3 \cdot \frac{4}{30} = 75.9 \]

Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box.

Solution: Convert to continuous intervals (add/subtract 0.5), use assumed mean method: a = 57, h = 3.

Mean: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \]

\[ \bar{x} = 57 + 3 \cdot \frac{25}{400} = 57 + 0.1875 \approx 57.19 \]

Q6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food.

Solution: Use assumed mean method, a = 225, h = 50.

Mean: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \]

\[ \bar{x} = 225 + 50 \cdot \frac{-7}{25} = 225 – 14 = 211 \]

Q7. To find out the concentration of SO₂ in the air (in ppm), the data was collected for 30 localities. Find the mean concentration of SO₂ in the air.

Solution: Use direct method (class marks small decimals).

Mean: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \]

\[ \bar{x} = \frac{2.96}{30} \approx 0.099\ \text{ppm} \]

Q8. A class teacher has the following absentee record of 40 students. Find the mean number of days a student was absent.

Solution: Direct method.

Mean: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \]

\[ \bar{x} = \frac{499}{40} \approx 12.48\ \text{days} \]

Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution: Assumed mean method, a = 70, h = 10.

Mean: \[ \bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} \]

\[ \bar{x} = 70 + 10 \cdot \frac{-2}{35} \approx 69.43\% \]