Maharashtra Board Class 10 Maths Chapter 6 Statistics Set 6.2 Solutions

Question 1. The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.

Solution:

Here, total frequency = \( \Sigma f_i \) = N = 1000
\[ \therefore \frac{N}{2} = \frac{1000}{2} = 500 \] Cumulative frequency which is just greater than (or equal) to 500 is 650.
\( \therefore \) The median class is 10 - 12.
Now, L = 10, f = 500, cf = 150, h = 2
\[ \therefore \text{Median} = L + \left[ \frac{\frac{N}{2} - \text{cf}}{\text{f}} \right] \text{h} \] \[ = 10 + \left( \frac{500-150}{500} \right) 2 \] \[ = 10 + 0.7 \times 2 \] \[ = 10 + 1.4 \] \[ = 11.4 \]
Answer: \( \therefore \) The median of the number of hours the workers work is 11.4 hours.
In simple words: To find the median, we first calculate the cumulative frequencies and determine the median class. Then, we apply the median formula using the lower limit (L), frequency (f), cumulative frequency (cf), and class width (h) of the median class.

🎯 Exam Tip: Remember to correctly identify the median class based on N/2 and the corresponding cumulative frequency. Carefully substitute values into the median formula to avoid calculation errors.

Question 2. The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.

Solution:

Here, total frequency = \( \Sigma f_i \) = N = 250
\[ \therefore \frac{N}{2} = \frac{250}{2} = 125 \] Cumulative frequency which is just greater than (or equal) to 125 is 153.
\( \therefore \) The median class is 150 - 200.
Now, L = 150, f = 90, cf = 63, h = 50
\[ \therefore \text{Median} = L + \left[ \frac{\frac{N}{2} - \text{cf}}{\text{f}} \right] \text{h} \] \[ = 150 + \left( \frac{125-63}{90} \right) 50 \] \[ = 150 + \left( \frac{62}{90} \right) 50 \] \[ = 150 + 34.4 \] \[ = 184.4 \approx 184 \]
Answer: \( \therefore \) The median of the given data is 184 mangoes (approx).
In simple words: We find the median by first arranging the data into a frequency distribution and calculating cumulative frequencies. The median class is identified using N/2, and then the median formula is applied to get the approximate median value.

🎯 Exam Tip: Pay attention to the class intervals and their corresponding frequencies. Ensure correct identification of the median class and accurate calculation of N/2, L, f, cf, and h for the median formula.

Question 3. The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.

Solution:

Here, total frequency = \( \Sigma f_i \) = N = 200
\[ \therefore \frac{N}{2} = \frac{200}{2} = 100 \] Cumulative frequency which is just greater than (or equal) to 100 is 184.
\( \therefore \) The median class is 74.5 - 79.5.
Now, L = 74.5, f = 85, cf = 99, h = 5
\[ \therefore \text{Median} = L + \left[ \frac{\frac{N}{2} - \text{cf}}{\text{f}} \right] \text{h} \] \[ = 74.5 + \left( \frac{100-99}{85} \right) 5 \] \[ = 74.5 + 0.059 \] \[ = 74.559 \approx 75 \]
Answer: \( \therefore \) The median of the given data is 75 km/hr (approx.).
In simple words: To find the median speed, we first convert the given classes into continuous classes. Then, we calculate cumulative frequencies to locate the median class using N/2. Finally, we apply the median formula for grouped data to get the approximate median speed.

🎯 Exam Tip: When given non-continuous classes, always convert them to continuous classes first by finding the average of the upper limit of one class and the lower limit of the next. This is crucial for accurate median calculation.

Question 4. The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

Solution:

Cumulative frequency which is just greater than (or equal) to 52.5 is 67.
\( \therefore \) The median class is 50 - 60.
Now, L = 50, f = 20, cf = 47, h = 10
\[ \therefore \text{Median} = L + \left[ \frac{\frac{N}{2} - \text{cf}}{\text{f}} \right] \text{h} \] \[ = 50 + \left( \frac{52.5-47}{20} \right) 10 \] \[ = 50 + \frac{5.5}{20} \times 10 \] \[ = 50 + 2.75 \] \[ = 52.75 \] \[ = 52.75 \times 1000 \] \[ = 52750 \text{ lamps} \]
Answer: \( \therefore \) The median of the productions is 52750 bulbs (approx.).
In simple words: To find the median production, we organize the data into a frequency table, calculate cumulative frequencies, and identify the median class. Using the median formula, we then compute the approximate median number of bulbs produced.

🎯 Exam Tip: Be careful with units. The "No. of bulbs produced" is in thousands, so the final median value of 52.75 needs to be multiplied by 1000 to get the actual number of bulbs, which is 52750.

Question 1. If the number of scores is odd, then the \( \left(\frac{n+1}{2}\right) \)th score is the median of the data. That is, the number of scores below as well as above \( K_{\frac{n+1}{2}} \) is \( \frac{n-1}{2} \). Verify the fact by taking n = 2m + 1. (Textbk pg. no. 139)
Solution:
Given that, n = 2m + 1
The sequence of the terms of scores can be 1, 2, 3, ...., 2m + 1
Here,
\[ \frac{n+1}{2} = \frac{2m+1+1}{2} = \frac{2m + 2}{2} = \frac{2(m+1)}{2} = m+1 \] The sequence of the terms of scores is 1, 2, 3, ......, m, m + 1, m + 2, ......, 2m + 1
Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1
i.e. to prove
number of terms from 1 to m = number of terms from m + 2 to 2m + 1 ...(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1, d = 1, \( t_{n1} \) = m
\( t_{n1} \) = a + (\( n_1 \) - 1) d
\[ \therefore m = 1 + (n_1 - 1)1 \] \[ \therefore m = 1 + n_1 - 1 \] \[ \therefore m = n_1 \] Consider the R.H.S. of equation (ii)
The sequence is an A.P. with a = m + 2, d = 1, \( t_{n2} \) = 2m + 1
\( t_{n2} \) = a + (\( n_2 \) - 1)d
\[ \therefore 2m + 1 = m + 2 + (n_2 - 1)1 \] \[ \therefore 2m + 1 = m + n_2 + 1 \] \[ \therefore m = n_2 \]
Answer: \( \therefore \) number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = \( \frac{n-1}{2} \).
\( \therefore \) m + 1 is the middle term if the number of scores is 2m + 1.
In simple words: For an odd number of scores (n = 2m+1), the median is the middle term, which is the (m+1)th term. This means there are 'm' terms before it and 'm' terms after it, confirming that the number of scores below and above the median is \( \frac{n-1}{2} \).

🎯 Exam Tip: When dealing with theoretical proofs in statistics, always clearly define 'n' (total number of scores) and 'm' (half of 'n-1') to set up the sequence correctly. Verifying both sides of the equality for terms below and above the median is essential.

Question 2. If the number of the scores is even, then the mean of the middle two terms is the median. This is because the number of terms below \( K_{\frac{n}{2}} \) and above \( K_{\frac{n+2}{2}} \) is equal, which is \( \frac{n-2}{2} \). Verify this by taking n = 2m. (Textbook pg. no. 139)
Solution:
Given that, n = 2m
The sequence of the terms of scores can be 1, 2, 3, ..., 2m.
Here \( \frac{n}{2} = \frac{2m}{2} = m \), and
\[ \frac{n+2}{2} = \frac{2m+2}{2} = \frac{2(m+1)}{2} = m+1 \] The sequence of the terms of scores is 1, 2, 3 ... m - 1, m, m + 1, m + 2,...., 2m
Thus, we have to prove that m and m + 1 are the middlemost terms if the number of scores is 2m.
i.e. to prove
number of terms from 1 to m - 1 = number of terms from m + 2 to 2m ...(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1, d = 1, \( t_{n1} \) = m - 1
\( t_{n1} \) = a + (\( n_1 \) - 1)d
\[ \therefore m - 1 = 1 + (n_1 - 1)1 \] \[ \therefore m - 1 = 1 + n_1 - 1 \] \[ \therefore n_1 = m - 1 \] Consider the R.H.S. of equation (i)
The sequence is an A.P. with a = m + 2, d = 1, \( t_{n2} \) = 2m
\( t_{n2} \) = a + (\( n_2 \) - 1) d
\[ \therefore 2m = m + 2 + (n_2 - 1)1 \] \[ \therefore 2m = m + 2 + n_2 - 1 \] \[ \therefore n_2 = m - 1 \]
Answer: \( \therefore \) number of terms from 1 to m - 1 = number of terms from m + 2 to 2m = m - 1 = \( \frac{n-2}{2} \).
\( \therefore \) m and m + 1 are the middlemost terms if the number of scores is 2m.
In simple words: For an even number of scores (n=2m), the median is the average of the two middle terms: the m-th term and the (m+1)-th term. This proof shows that the number of terms before the m-th term and after the (m+1)-th term is equal, which is \( \frac{n-2}{2} \), confirming their central position.

🎯 Exam Tip: When proving median concepts for even 'n', precisely identifying the two middle terms (n/2 and n/2+1) and then verifying the number of terms on either side of these central terms is crucial for a complete and accurate answer.