CBSE Class 10 Mathematics Statistics Assignment Set 16

Very Short Answer Type Questions

Question 1. The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean?
Answer:
If every observation is decreased by 5, the mean also decreases by 5.
New mean = 20 - 5 = 15.
In simple words: Subtracting a constant from every observation subtracts it from the mean.

Teacher's Note:
1. Read what is given and what is asked, then write the formula you need before substituting.
2. Check the final value by putting it back in the condition of the question.

Question 2. The marks obtained by 10 students in physics out of 40 are 24, 27, 29, 34, 32, 19, 26, 35, 18, 21. Compute the mean of the marks.
Answer:
Sum of marks = 24 + 27 + 29 + 34 + 32 + 19 + 26 + 35 + 18 + 21 = 265.
Mean \( = \frac{265}{10} = 26.5 \).
In simple words: Add all the marks and divide by 10.

Teacher's Note:
1. Read what is given and what is asked, then write the formula you need before substituting.
2. Check the final value by putting it back in the condition of the question.

Question 3. Find the class marks of the classes 10 - 25 and 35 - 55.
Answer:
Class mark \( = \frac{\text{lower limit} + \text{upper limit}}{2} \).
For 10 - 25: \( \frac{10 + 25}{2} = 17.5 \). For 35 - 55: \( \frac{35 + 55}{2} = 45 \).
In simple words: The class mark is the middle value of the class.

Teacher's Note:
1. Read what is given and what is asked, then write the formula you need before substituting.
2. Check the final value by putting it back in the condition of the question.

Question 4. Find the class marks of the classes 20 - 50 and 35 - 60.
Answer:
For 20 - 50: \( \frac{20 + 50}{2} = 35 \). For 35 - 60: \( \frac{35 + 60}{2} = 47.5 \).
In simple words: Average the two limits of each class.

Teacher's Note:
1. Read what is given and what is asked, then write the formula you need before substituting.
2. Check the final value by putting it back in the condition of the question.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

Question 7. For what value of x is the mode of the following data 8?
4, 5, 6, 8, 5, 4, 8, 5, 6, x, 8, 4
Answer:
The frequencies of the known values are: 4 occurs 3 times, 5 occurs 3 times, 6 occurs 2 times, 8 occurs 3 times.
For 8 to be the only mode, its frequency must exceed the others, so the missing observation must be 8. Then 8 occurs 4 times.
Hence x = 8.
In simple words: Add x to the value that should become the mode, so that it occurs most often.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.

Question 9. Find the median of the first 100 natural numbers.
Answer:
The numbers 1, 2, ..., 100 are in order and n = 100 is even.
Median \( = \frac{50^{th} + 51^{st}}{2} = \frac{50 + 51}{2} = 50.5 \).
In simple words: With an even number of terms, average the two middle ones.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

Question 10. If the median of the observations 9, 13, 15, 20, x + 2, x + 4, 33, 45, 56, 61 arranged in ascending order is 24, find the value of x.
Answer:
There are 10 observations, so the median is the mean of the 5th and 6th observations: \( \frac{(x + 2) + (x + 4)}{2} = 24 \).
\( x + 3 = 24 \Rightarrow x = 21 \).
In simple words: The two middle terms are x + 2 and x + 4.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

Question 12. Find the median of the data, using an empirical relation, when it is given that Mode = 12.4 and Mean = 10.5.
Answer:
Empirical relation: Mode = 3 Median - 2 Mean.
\( 12.4 = 3 \times \text{Median} - 2(10.5) \Rightarrow 3 \times \text{Median} = 33.4 \).
Median \( = \frac{33.4}{3} \approx 11.13 \).
In simple words: Use Mode = 3 Median - 2 Mean and solve for the median.

Teacher's Note:
1. Read what is given and what is asked, then write the formula you need before substituting.
2. Check the final value by putting it back in the condition of the question.

Short Answer Type Questions

Question 13. The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m - 1 and median q. Find the values of p and q.
Answer:
\( m = \frac{1 + 7 + 5 + 3 + 4 + 4}{6} = \frac{24}{6} = 4 \), so the mean of the second set is \( m - 1 = 3 \).
\( \frac{3 + 2 + 4 + 2 + 3 + 3 + p}{7} = 3 \Rightarrow 17 + p = 21 \Rightarrow p = 4 \).
Arranging 2, 2, 3, 3, 3, 4, 4 in order, the middle (4th) value is 3, so q = 3.
In simple words: Find m first, then use the mean to get p and the arranged list to get q.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

Question 14. If the mean of n observations \( ax_1, ax_2, ax_3, \ldots, ax_n \) is \( a\bar{x} \), show that \( (ax_1 - a\bar{x}) + (ax_2 - a\bar{x}) + \cdots + (ax_n - a\bar{x}) = 0 \).
Answer:
The sum of the deviations is \( \sum (ax_i - a\bar{x}) = a\sum x_i - n \cdot a\bar{x} \).
Since \( \bar{x} = \frac{\sum x_i}{n} \), we have \( \sum x_i = n\bar{x} \), so the sum \( = a \cdot n\bar{x} - n a\bar{x} = 0 \). Hence shown.
In simple words: The deviations from the mean always add up to zero.

Teacher's Note:
1. Read what is given and what is asked, then write the formula you need before substituting.
2. Check the final value by putting it back in the condition of the question.

Question 15. If \( x_1, x_2, \ldots, x_n \) are n values of a variable X such that \( \sum_{i=1}^{n}(x_i - 2) = 110 \) and \( \sum_{i=1}^{n}(x_i - 5) = 20 \), find the value of n and the mean.
Answer:
\( \sum (x_i - 2) - \sum (x_i - 5) = 110 - 20 \Rightarrow \sum (5 - 2) = 90 \Rightarrow 3n = 90 \Rightarrow n = 30 \).
\( \sum x_i - 2n = 110 \Rightarrow \sum x_i = 110 + 60 = 170 \).
Mean \( = \frac{170}{30} = \frac{17}{3} \approx 5.67 \).
In simple words: Subtract the two sums to find n, then find the sum of the observations.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

Question 16. The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is 100. It was later found that it is 110, not 100. Find the true mean and median.
Answer:
Sum of the observations = \( 100 \times 50 = 5000 \). Corrected sum = \( 5000 - 100 + 110 = 5010 \).
True mean \( = \frac{5010}{100} = 50.1 \).
The median depends on the middle observations only. The largest observation is still the largest (110 instead of 100), so the order does not change and the true median remains 52.
In simple words: Correct the total for the mean; the median is not affected by the change in the largest value.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.