Maharashtra Board Class 10 Maths Chapter 6 Statistics Set 6.3 Solutions

Question 1. The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.

Answer: Solution:

Here, the maximum frequency is 80. \( \therefore \) The modal class is 4 - 5. L = lower class limit of the modal class = 4 h = class interval of the modal class = 1 \( f_1 \) = frequency of the modal class = 80 \( f_0 \) = frequency of the class preceding the modal class = 70 \( f_2 \) = frequency of the class succeeding the modal class = 60
\( \therefore \) Mode \( = L + \left( \frac{f_1-f_0}{2f_1-f_0-f_2} \right) h \)
\( = 4 + \left( \frac{80-70}{2(80)-70-60} \right) 1 \)
\( = 4 + \left( \frac{10}{160-130} \right) 1 \)
\( = 4 + \frac{10}{30} \)
\( = 4 + 0.33 \)
\( = 4.33 \) \( \therefore \) The mode of the fat content is 4.33%.
In simple words: The mode of fat content is calculated using the formula for mode of grouped data, identifying the modal class, its lower limit, frequencies of modal, preceding, and succeeding classes, and class interval. The highest frequency determines the modal class.

🎯 Exam Tip: Remember to correctly identify \( L, h, f_1, f_0 \), and \( f_2 \) from the frequency table to avoid calculation errors in the mode formula.

Question 2. Electricity used by some families is shown in the following table. Find the mode of use of electricity.

Answer: Solution: Mean \( = \overline{X} = \frac{\sum_{i=1}^{N} f_i x_i}{\Sigma f_i} = \frac{69500}{50} = 1390 \)

Here, the maximum frequency is 100.
\( \therefore \) The modal class is 60 - 80. L = lower class limit of the modal class = 60 h = class interval of the modal class = 20 \( f_1 \) = frequency of the modal class = 100 \( f_0 \) = frequency of the class preceding the modal class = 70 \( f_2 \) = frequency of the class succeeding the modal class = 80
\( \therefore \) Mode \( = L + \left( \frac{f_1-f_0}{2f_1-f_0-f_2} \right) h \)
\( = 60 + \left( \frac{100-70}{2(100)-70-80} \right) 20 \)
\( = 60 + \left( \frac{30}{200-150} \right) 20 \)
\( = 60 + \frac{30}{50} \times 20 \)
\( = 60 + \frac{600}{50} \)
\( = 60 + 12 \)
\( = 72 \) \( \therefore \) The mode of use of electricity is 72 units.
In simple words: The modal electricity usage is found by identifying the class with the highest frequency (modal class) and applying the mode formula for grouped data. This calculation determines the electricity consumption value that occurs most frequently among the families.

🎯 Exam Tip: Pay close attention to the class intervals and corresponding frequencies when setting up the mode formula, especially for \( L \) and \( h \).

Question 3. Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.

Answer: Solution:

Here, the maximum frequency is 35.
\( \therefore \) The modal class is 9 - 11. L = lower class limit of the modal class = 9 h = class interval of the modal class = 2 \( f_1 \) = frequency of the modal class = 35 \( f_0 \) = frequency of the class preceding the modal class = 20 \( f_2 \) = frequency of the class succeeding the modal class = 18
\( \therefore \) Mode \( = L + \left( \frac{f_1-f_0}{2f_1-f_0-f_2} \right) h \)
\( = 9 + \left( \frac{35-20}{2(35)-20-18} \right) 2 \)
\( = 9 + \left( \frac{15}{70-38} \right) 2 \)
\( = 9 + \left( \frac{15}{32} \right) 2 \)
\( = 9 + 0.9375 \)
\( = 9.9375 \approx 9.94 \) \( \therefore \) The mode of the supply of milk is 9.94 litres (approx.).
In simple words: To find the mode of milk supply, identify the class with the highest number of hotels (modal class). Then, apply the formula for the mode of grouped data, using the values for the lower limit, class interval, and frequencies of the modal, preceding, and succeeding classes.

🎯 Exam Tip: Double-check the arithmetic, especially in the denominator \( (2f_1-f_0-f_2) \), as small errors can significantly impact the final mode value.

Question 4. The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.

Answer: Solution:

Here, the maximum frequency is 50. The modal class is 9.5 - 14.5. L = lower class limit of the modal class = 9.5 h = class interval of the modal class = 5 \( f_1 \) = frequency of the modal class = 50 \( f_0 \) = frequency of the class preceding the modal class = 32 \( f_2 \) = frequency of the class succeeding the modal class = 36
\( \therefore \) Mode \( = L + \left( \frac{f_1-f_0}{2f_1-f_0-f_2} \right) h \)
\( = 9.5 + \left( \frac{50-32}{2(50)-32-36} \right) 5 \)
\( = 9.5 + \left( \frac{18}{100-68} \right) 5 \)
\( = 9.5 + \left( \frac{18}{32} \right) 5 \)
\( = 9.5 + 2.8125 \)
\( = 12.3125 \approx 12.31 \) \( \therefore \) The mode of the ages of the patients is 12.31 years (approx.).
In simple words: To find the mode of patient ages, first convert the classes to continuous form. Then, identify the class with the highest frequency (modal class) and use the mode formula for grouped data, including the lower limit, class interval, and frequencies to calculate the most common age.

🎯 Exam Tip: When given non-continuous classes like "Less than 5" or "5-9", remember to convert them into continuous class intervals (e.g., 0-4.5, 4.5-9.5) before applying the mode formula. This is a common pitfall.