GSEB Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1. 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 of plants per house. Which method did you use for finding the mean and why?
Answer: A survey was performed by some students for an environment awareness program. They gathered data about the number of plants in 20 houses in one area. We need to calculate the average number of plants per house.

Answer: Using the direct method, the mean is calculated as:
\( \bar{x} = \frac { \Sigma f_i x_i }{ \Sigma f_i } = \frac { 162 }{ 20 } = 8.1 \) plants.
We used the direct method for finding the mean because the numbers for \( x_i \) and \( f_i \) are small, making calculations simple.
In simple words: Students collected data on plants in 20 houses. We need to find the average plants per house. The direct method was used because the numbers were small and simple, giving a mean of 8.1 plants.

Exam Tip: When numerical values in the dataset are small, the direct method is the most straightforward and accurate way to determine the mean.

Question 2. 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.
Answer: The distribution of daily earnings for 50 factory workers is given. We need to determine the average daily earnings of these workers using an appropriate calculation method.

Answer: We are using the step-deviation approach, with assumed mean \( a = 150 \) and class size \( h = 20 \).
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = a + \left( \frac { \Sigma f_i u_i }{ \Sigma f_i } \right) \times h \)
\( \bar{x} = 150 + \left( \frac { -12 }{ 50 } \right) \times 20 \)
\( \bar{x} = 150 - 4.8 \)
\( \bar{x} = 145.20 \)
Based on our calculations, the average daily earnings for the factory workers come out to be Rs. 145.20.
In simple words: We need to find the average daily pay for 50 factory workers. Using the step-deviation method, the average daily wage is Rs. 145.20.

Exam Tip: The step-deviation method simplifies calculations when class intervals are equal and values are large, making it efficient for grouped data.

Question 3. 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.
Answer: This distribution displays the daily pocket money of children in a certain area. The average pocket money is already given as Rs. 18. Our task is to find the unknown frequency, f.

Answer: We apply the direct method for calculations. The mean \( \bar{x} \) is given by:
\( \bar{x} = \frac { \Sigma f_i x_i }{ \Sigma f_i } \)
Substituting the given mean and calculated sums:
\( 18 = \frac { 20f + 752 }{ f + 44 } \)

\( \implies \) \( 18(f + 44) = 20f + 752 \)

\( \implies \) \( 18f + 792 = 20f + 752 \)

\( \implies \) \( 792 - 752 = 20f - 18f \)

\( \implies \) \( 40 = 2f \)

\( \implies \) \( f = \frac { 40 }{ 2 } \)

\( \implies \) \( f = 20 \)
Therefore, the value for the missing frequency, f, is 20.
In simple words: We have a list of children's daily pocket money with an average of Rs. 18. We need to find a missing number, 'f', in the list. By using the direct method and solving the equation, we find that 'f' is 20.

Exam Tip: When a mean is given along with a missing frequency, set up the mean formula as an equation and solve for the unknown variable.

Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heartbeat per minute for these women, choosing a suitable method.
Answer: A doctor at a hospital checked thirty women, noting their heartbeats per minute, and summarized the data. We need to calculate the average heartbeat rate for these women, selecting an appropriate method.

Answer: We are using the step-deviation approach, with assumed mean \( a = 75.5 \) and class size \( h = 3 \).
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = a + \left( \frac { \Sigma f_i u_i }{ \Sigma f_i } \right) \times h \)
\( \bar{x} = 75.5 + \left( \frac { 4 }{ 30 } \right) \times 3 \)
\( \bar{x} = 75.5 + 0.4 \)
\( \bar{x} = 75.9 \)
After calculations, the average heartbeat per minute for these women is 75.9.
In simple words: A doctor recorded heartbeats for 30 women. We need to find their average heartbeat per minute. Using the step-deviation method, the average heartbeat rate is 75.9.

Exam Tip: For grouped data with equal class widths, the step-deviation method often makes calculations simpler and reduces the chance of errors.

Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes. Which method of finding the mean did you choose?
Answer: In a market, fruit sellers sold mangoes in boxes, with different amounts in each box. The table shows how mangoes were divided across boxes. We need to determine the average number of mangoes per box and explain our chosen method.

Answer: We applied the step-deviation method, with assumed mean \( a = 57 \) and class size \( h = 3 \).
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = a + \left( \frac { \Sigma f_i u_i }{ \Sigma f_i } \right) \times h \)
\( \bar{x} = 57 + \left( \frac { 25 }{ 400 } \right) \times 3 \)
\( \bar{x} = 57 + 0.19 \)
\( \bar{x} = 57.19 \)
We chose this approach because the frequency and class mark numbers were quite big, which helps to simplify calculations. The average number of mangoes per box is 57.19.
In simple words: Fruit sellers had mangoes in boxes. We need to find the average number of mangoes per box. The step-deviation method was chosen because the numbers were large, making it easier to calculate. The mean is 57.19 mangoes.

Exam Tip: The step-deviation method is particularly useful when the numerical values of the class marks and frequencies are large, simplifying the arithmetic significantly.

Question 6. The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Answer: The table below presents the daily amount spent on food by 25 families in one area. We need to calculate the average daily spending on food using a suitable technique.

Answer: We employed the step-deviation approach, with assumed mean \( a = 225 \) and class size \( h = 50 \).
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = a + \left( \frac { \Sigma f_i u_i }{ \Sigma f_i } \right) \times h \)
\( \bar{x} = 225 + \left( \frac { -7 }{ 25 } \right) \times 50 \)
\( \bar{x} = 225 - 14 \)
\( \bar{x} = 211 \)
Based on our calculations, the average daily amount spent on food is Rs. 211.
In simple words: This table shows how much 25 families spend on food daily. We need to find the average daily food cost. Using the step-deviation method, the average daily expenditure on food is Rs. 211.

Exam Tip: When dealing with larger numbers or wider class intervals, the step-deviation method helps to simplify the computations and maintain accuracy.

Question 7. To find out the concentration of SO2 in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below. Find the mean concentration of SO2 in the air.
Answer: To determine the level of SO2 in the air (measured in parts per million), information was gathered from 30 different places in a city and is shown below. We need to calculate the average concentration of SO2 in the atmosphere.

Answer: We are utilizing the step-deviation method, with assumed mean \( a = 0.14 \) and class size \( h = 0.04 \).
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = a + \left( \frac { \Sigma f_i u_i }{ \Sigma f_i } \right) \times h \)
\( \bar{x} = 0.14 + \left( \frac { -31 }{ 30 } \right) \times 0.04 \)
\( \bar{x} = 0.14 - 0.041 \)
\( \bar{x} = 0.099 \) ppm
As a result, the average concentration of SO2 in the air is determined to be 0.099 ppm.
In simple words: Data was collected from 30 city areas to find the air's SO2 concentration. We need to find the average SO2 level. Using the step-deviation method, the mean SO2 concentration is 0.099 ppm.

Exam Tip: For continuous grouped data like concentration levels, using appropriate statistical methods ensures accurate representation of the mean value.

Question 8. A class teacher has the following absent students of a class for the whole term. Find the mean number of days a student was absent.
Answer: A teacher recorded the days students were absent throughout the term for a class. We need to calculate the average number of days a student remained absent.

Answer: We are using the direct approach for this calculation.
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = \frac { \Sigma f_i x_i }{ \Sigma f_i } \)
\( \bar{x} = \frac { 499 }{ 40 } \)
\( \bar{x} = 12.47 \)
Based on the data, the average number of days a student was absent comes out to be 12.47.
In simple words: A teacher recorded student absences. We need to find the average number of absent days per student. Using the direct method, the average absence is 12.47 days.

Exam Tip: The direct method is suitable when the values of \( f_i \) and \( x_i \) are relatively small, simplifying the summation process for the mean.

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Answer: The table provided shows the literacy percentage for 35 different cities. We need to calculate the average literacy rate.

Answer: We are applying the step-deviation method, with assumed mean \( a = 70 \) and class size \( h = 10 \).
The mean \( \bar{x} \) is calculated as:
\( \bar{x} = a + \left( \frac { \Sigma f_i u_i }{ \Sigma f_i } \right) \times h \)
\( \bar{x} = 70 + \left( \frac { -2 }{ 35 } \right) \times 10 \)
\( \bar{x} = 70 - \frac { 20 }{ 35 } \)
\( \bar{x} = 70 - \frac { 4 }{ 7 } \)
\( \bar{x} = 70 - 0.57 \)
\( \bar{x} = 69.43\% \)
After completing the calculations, the average literacy rate for these cities is found to be 69.43%.
In simple words: This table shows how many people can read and write in 35 cities. We need to find the average literacy rate. Using the step-deviation method, the average is 69.43%.

Exam Tip: When working with percentages or grouped data, ensure the correct class marks are derived and the step-deviation method is applied accurately for precise mean calculation.