Maharashtra Board Class 10 Maths Chapter 6 Statistics Set 6.1 Solutions

Question 1. The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.

Answer: Solution:

\[Mean = \bar{X} = \frac{\sum_{i=1}^{N}f_ix_i}{\sum f_i} = \frac{218}{50} = 4.36\] Therefore, The mean of the time spent by the students for their studies is 4.36 hours.
In simple words: To find the average study time, we multiply the midpoint of each time interval by the number of students in that interval, sum these products, and then divide by the total number of students. This gives us the mean study time using the direct method.

🎯 Exam Tip: Always clearly define class marks and show the calculation of \(f_ix_i\) for each class. Ensure the summation of \(f_i\) and \(f_ix_i\) is accurate for a correct mean calculation.

Question 2. In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by 'assumed mean' method.

Answer: Solution: Let us take the assumed mean (A) = 550

Here, \(\sum f_id_i = -12000\), \(\sum f_i = 420\)
\(\bar{d} = \frac{\sum f_id_i}{\sum f_i} = \frac{-12000}{420} = -28.57\)
\(Mean = \bar{X} = A + \bar{d}\)
\(= 550 + (-28.57)\)
\(= 521.43\)
Therefore, The mean of the toll paid by the drivers is Rs. 521.43.
In simple words: The assumed mean method simplifies calculating the mean by picking an arbitrary class mark (assumed mean), finding deviations from it, and then adjusting the assumed mean by the average of these deviations. This makes calculations easier for large datasets.

🎯 Exam Tip: Choose an assumed mean (A) from the middle of the class marks to minimize deviation values, making calculations simpler and reducing error potential.

Question 3. A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.

Answer: Solution:

Here, \(\sum f_ix_i = 141\), \(\sum f_i = 50\)
\(Mean = \bar{X} = \frac{\sum_{i=1}^{N}f_ix_i}{\sum f_i} = \frac{141}{50} = 2.82\)
Therefore, The mean of the milk sold is 2.82 litres.
In simple words: To find the average milk sold, we multiply the midpoint of each litre range by the number of customers who bought that much, sum these products, and then divide by the total number of customers. This directly calculates the mean milk sold.

🎯 Exam Tip: Ensure that the class mark for each interval is correctly calculated as the midpoint. This is a common source of error in direct method mean calculations.

Question 4. A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by 'assumed mean' method.

Answer: Solution: Let us take the assumed mean (A) = 37.5

Here, \(\sum f_id_i = -175\), \(\sum f_i = 80\)
\(\bar{d} = \frac{\sum f_id_i}{\sum f_i} = \frac{-175}{80} = -2.19\)
\(Mean = \bar{X} = A + \bar{d}\)
\(= 37.5 + (-2.19)\)
\(= 35.31\)
\(= (\text{35.31} \times 1000)\)
\(= 35310\)
Therefore, The mean of the production of oranges is Rs. 35310.
In simple words: The mean production is calculated by assuming an average value, finding how much each class deviates from this assumption, and then adjusting the assumed mean by the weighted average of these deviations. This helps find the true average production more efficiently.

🎯 Exam Tip: Remember to multiply the final mean deviation (\(\bar{d}\)) by the appropriate unit (e.g., 1000 for 'Thousand rupees') if the initial data was given in such units.

Question 5. A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of he funds by 'step deviation' method.

Answer: Solution: Here, we take A = 1250 and g = 500

Here, \(\sum f_iu_i = -63\), \(\sum f_i = 120\), g = 500
\(\bar{u} = \frac{\sum f_iu_i}{\sum f_i} = \frac{-63}{120} = -0.525\)
\(Mean = \bar{X} = A + \bar{u}g = 1250 + (-0.525 \times 500)\)
\(= 1250 - 262.5\)
\(= 987.5\)
Therefore, The mean of the funds collected is Rs. 987.5.
In simple words: The step deviation method simplifies mean calculation for grouped data by converting deviations into smaller, easier-to-handle units (steps) before averaging. This is especially useful when class intervals are uniform and deviation values are large.

🎯 Exam Tip: When using the step deviation method, ensure the class width (g) is consistently applied to all deviations. A common mistake is using an incorrect 'g' or miscalculating \(u_i\).

Question 6. The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.

Answer: Solution: Here, we take A = 2500 and g = 1000.

Here, \(\sum f_iu_i = 85\), \(\sum f_i = 150\)
\(\bar{u} = \frac{\sum f_iu_i}{\sum f_i} = \frac{85}{150} = 0.57\)
\(Mean = \bar{X} = A + \bar{u}g\)
\(= 2500 + 0.57 (1000)\)
\(= 2500 + 570 = 3070\)
Therefore, The mean of the weekly wages is Rs. 3070.
In simple words: The mean weekly wage is found by taking an assumed mean, calculating deviations, scaling those deviations by the class width, averaging the scaled deviations, and then adding this average back to the assumed mean. This streamlines the calculation for large numbers.

🎯 Exam Tip: Pay close attention to the formula for the step deviation method: \(Mean = A + \bar{u}g\). A common mistake is forgetting to multiply \(\bar{u}\) by the class width 'g'.

Question 1. The daily sale of 100 vegetable vendors is given in the following table. Find the mean of the sale by direct method. (Textbook pg. no. 133 and 134)

Answer: Solution:

\[Mean = \bar{X} = \frac{\sum_{i=1}^{N}f_ix_i}{\sum f_i} = \frac{215000}{100} = 2150\] The mean of the sale is 2150.
In simple words: To find the average daily sale, we multiply the midpoint of each sales range by the number of vendors in that range, sum these products, and then divide by the total number of vendors. This directly gives us the mean sale amount.

🎯 Exam Tip: Always double-check the class marks and the multiplication of \(f_i\) and \(x_i\) when using the direct method to avoid arithmetic errors.

Question 2. The amount invested in health insurance by 100 families is given in the following frequency table. Find the mean of investments using direct method and assumed mean method. Check whether the mean found by the two methods is the same as calculated by step deviation method (Ans: Rs. 2140). (Textbook pg. no. 135 and 136)

Answer: Solution: Direct method:

\[Mean = \bar{X} = \frac{\sum_{i=1}^{N}f_ix_i}{\sum f_i} = \frac{214000}{100} = 2140\] Therefore, The mean of investments in health insurance is Rs. 2140.
Assumed mean method:

\(\bar{d} = \frac{\sum f_id_i}{\sum f_i} = \frac{-6000}{100} = -60\)
\(Mean = \bar{X} = A + \bar{d}\)
\(= 2200 + (-60) = 2140\)
Therefore, The mean of investments in health insurance is Rs. 2140.
Therefore, Mean found by direct method and assumed mean method is the same as calculated by step deviation method.
In simple words: This question demonstrates that the mean calculated using the direct method and the assumed mean method yield the same result. Both methods determine the average investment by families, with the assumed mean method often simplifying calculations by using deviations from a chosen central value.

🎯 Exam Tip: When asked to compare different methods (direct, assumed mean, step deviation), ensure you present the full calculations for each and explicitly state the conclusion about their equality if it's found to be true.

Question 3. The following table shows the funds collected by 50 students for flood affected people. Find the mean of the funds.

Answer: Solution: If the number of scores in two consecutive classes is very low, it is convenient to club them. So, in the above example, we club the classes 0 - 500, 500 - 1000 and 2000 - 2500, 2500 - 3000. Now the new table is as follows

(i) Solve by direct method.
(ii) Verily that the mean calculated by assumed mean method is the same.
(iii) Find the mean in the above example by taking A = 1750. (Textbook pg. no. 137)
Solution:
(i) Direct method:

\[Mean = \bar{X} = \frac{\sum_{i=1}^{N}f_ix_i}{\sum f_i} = \frac{69500}{50} = 1390\] Therefore, The mean of the funds is Rs. 1390.
(ii) Assumed mean method: Here, A = 1250

\(\bar{d} = \frac{\sum f_id_i}{\sum f_i} = \frac{7000}{50} = 140\)
\(Mean = \bar{X} = A + \bar{d}\)
\(= 1250 + 140\)
\(= 1390\)
Therefore, The mean calculated by assumed mean method is the same.
(iii) Step deviation method: Here, we take A = 1750 and g = 250

\(\bar{u} = \frac{\sum f_iu_i}{\sum f_i} = \frac{-72}{50} = -1.44\)
\(Mean = \bar{X} = A + \bar{u}g\)
\(= 1750 + (-1.44) \times 250\)
\(= 1750 - 360\)
\(= 1390\)
Therefore, The mean of the funds is Rs. 1390.
In simple words: This problem illustrates how to find the mean using three different methods: direct, assumed mean, and step deviation. The initial data is grouped to simplify calculations, and all three methods ultimately confirm the same average fund collected, demonstrating their equivalence in finding the mean of grouped data.

🎯 Exam Tip: When class frequencies are very low at the extremes, clubbing classes is a valid strategy to simplify calculations, but ensure the new class intervals are well-defined and their class marks accurately reflect the combined range.