CBSE Class 10 Mathematics Statistics Assignment Set 03

Question 1. Marks obtained by 70 students are given below:

Find the median. (Ans: 50)
Answer:
First, arrange the marks in ascending order and construct the cumulative frequency table:

Total number of observations, \(N = 70\)
\(\frac{N}{2} = \frac{70}{2} = 35\)
The cumulative frequency just greater than or equal to 35 is 38, and the corresponding value of marks is 50.
Hence, the median is 50.
In simple words: Order the data, find cumulative frequencies, look at half the total observations, and pick the corresponding mark value.

Exam Tip: Always arrange discrete data points in ascending order of values before calculating cumulative frequencies.

Question 2. The sum of deviations of a set of values \(x_1, x_2, x_3, \dots, x_n\) measured from 50 is -10 and the sum of deviations of the values from 46 is 70. Find the value of n and the mean. (Ans: 20, 49.5)
Answer:
We are given:
\(\sum_{i=1}^{n} (x_i - 50) = -10\)
\(\sum_{i=1}^{n} x_i - 50n = -10 \quad \text{--- (1)}\)
And:
\(\sum_{i=1}^{n} (x_i - 46) = 70\)
\(\sum_{i=1}^{n} x_i - 46n = 70 \quad \text{--- (2)}\)
Subtracting equation (2) from equation (1):
\((-50n) - (-46n) = -10 - 70\)
\(-4n = -80\)
\(n = 20\)
Substitute \(n = 20\) in equation (1):
\(\sum_{i=1}^{n} x_i - 50(20) = -10\)
\(\sum_{i=1}^{n} x_i - 1000 = -10\)
\(\sum_{i=1}^{n} x_i = 990\)
Now, calculate the mean:
\(\text{Mean } (\overline{x}) = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{990}{20} = 49.5\)
Hence, \(n = 20\) and mean = 49.5.
In simple words: Set up simultaneous linear equations from the given deviation sums to solve for n and the total sum of observations.

Exam Tip: Express deviation sums as \(\sum x_i - a \cdot n = \text{Sum}\) to easily eliminate \(\sum x_i\) by subtraction.

Question 3. Prove that \(\sum_{i=1}^{n} (x_i - \overline{x}) = 0\)
Answer:
To prove that the algebraic sum of deviations of a set of observations from their mean is zero:
We know that mean \(\overline{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\)
Multiplying both sides by \(n\):
\(n\overline{x} = \sum_{i=1}^{n} x_i\)
Now consider the sum of deviations:
\(\sum_{i=1}^{n} (x_i - \overline{x}) = (x_1 - \overline{x}) + (x_2 - \overline{x}) + \dots + (x_n - \overline{x})\)
\(= (x_1 + x_2 + \dots + x_n) - (\overline{x} + \overline{x} + \dots + \overline{x} \text{ [n times]})\)
\(= \sum_{i=1}^{n} x_i - n\overline{x}\)
Since \(\sum_{i=1}^{n} x_i = n\overline{x}\), we substitute this:
\(= n\overline{x} - n\overline{x} = 0\)
Hence proved, \(\sum_{i=1}^{n} (x_i - \overline{x}) = 0\).
In simple words: Breaking the summation into two parts reveals that the sum of all data points equals n times the mean, resulting in zero difference.

Exam Tip: This is a standard theoretical proof frequently asked in exams. Memorize the substitution step \(\sum x_i = n\overline{x}\).

Question 4. Compute the median from the following data:

(Ans: 153.8)
Answer:
Since mid-values are given, we first find the class intervals. The class size \(h = 125 - 115 = 10\).
Lower limit = \(\text{Midvalue} - \frac{h}{2}\)
Upper limit = \(\text{Midvalue} + \frac{h}{2}\)

Here, \(N = \sum f_i = 390\), so \(\frac{N}{2} = \frac{390}{2} = 195\)
The cumulative frequency just greater than 195 is 267, corresponding to the median class \(150 - 160\).
Lower limit \(L = 150\), frequency of median class \(f = 116\), cumulative frequency of preceding class \(cf = 151\), class size \(h = 10\).
\(\text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
\(\text{Median} = 150 + \left( \frac{195 - 151}{116} \right) \times 10\)
\(\text{Median} = 150 + \frac{44 \times 10}{116} = 150 + \frac{440}{116} = 150 + 3.793 = 153.79 \approx 153.8\)
In simple words: Convert mid-values to continuous class intervals first, find the median class using N/2, and apply the standard grouped median formula.

Exam Tip: When mid-values are given, remember to subtract and add half the class interval (\(h/2\)) to determine correct class boundaries.

Question 5. The mean of 'n' observation is \(\overline{x}\), if the first term is increased by 1, second by 2 and so on. What will be the new mean. (Ans: \(\overline{x} + \frac{n+1}{2}\))
Answer:
Let the original observations be \(x_1, x_2, x_3, \dots, x_n\) with mean \(\overline{x}\).
The new observations after the respective increases are:
\(x_1 + 1, \, x_2 + 2, \, x_3 + 3, \, \dots, \, x_n + n\)
The new mean \(\overline{x}_{\text{new}}\) is given by:
\(\overline{x}_{\text{new}} = \frac{(x_1 + 1) + (x_2 + 2) + (x_3 + 3) + \dots + (x_n + n)}{n}\)
Rearranging the terms:
\(\overline{x}_{\text{new}} = \frac{(x_1 + x_2 + \dots + x_n) + (1 + 2 + 3 + \dots + n)}{n}\)
\(\overline{x}_{\text{new}} = \frac{\sum_{i=1}^{n} x_i}{n} + \frac{\frac{n(n+1)}{2}}{n}\)
Since \(\frac{\sum x_i}{n} = \overline{x}\):
\(\overline{x}_{\text{new}} = \overline{x} + \frac{n(n+1)}{2n} = \overline{x} + \frac{n+1}{2}\)
In simple words: Separate the original terms from the incremental values, sum the AP series of increments, and divide by n.

Exam Tip: Use the formula for the sum of first n natural numbers \(\frac{n(n+1)}{2}\) to simplify variable increments quickly.

Question 6. In a frequency distribution mode is 7.88, mean is 8.32 find the median. (Ans: 8.17)
Answer:
Using the empirical relationship between mean, median, and mode:
\(\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}\)
Substitute the given values into the formula:
\(7.88 = 3 \times \text{Median} - 2 \times 8.32\)
\(7.88 = 3 \times \text{Median} - 16.64\)
\(3 \times \text{Median} = 7.88 + 16.64\)
\(3 \times \text{Median} = 24.52\)
\(\text{Median} = \frac{24.52}{3} = 8.1733 \approx 8.17\)
In simple words: Apply the standard statistical formula connecting mode, median, and mean to find the missing value.

Exam Tip: Always memorize the empirical formula: \(\text{Mode} = 3(\text{Median}) - 2(\text{Mean})\).

Question 7. The mode of a distribution is 55 & the modal class is 45-60 and the frequency preceding the modal class is 5 and the frequency after the modal class is 10. Find the frequency of the modal class. (Ans: 15)
Answer:
Given parameters:
Mode (\(Z\)) = 55
Modal class = \(45 - 60 \implies L = 45\), class size \(h = 60 - 45 = 15\)
Frequency of preceding class (\(f_1\)) = 5
Frequency of succeeding class (\(f_2\)) = 10
Let the frequency of the modal class be \(f\).
The formula for mode is:
\(\text{Mode} = L + \left( \frac{f - f_1}{2f - f_1 - f_2} \right) \times h\)
Substituting the given values:
\(55 = 45 + \left( \frac{f - 5}{2f - 5 - 10} \right) \times 15\)
\(55 - 45 = \left( \frac{f - 5}{2f - 15} \right) \times 15\)
\(10 = \frac{15(f - 5)}{2f - 15}\)
Divide both sides by 5:
\(2 = \frac{3(f - 5)}{2f - 15}\)
\(2(2f - 15) = 3(f - 5)\)
\(4f - 30 = 3f - 15\)
\(4f - 3f = 30 - 15\)
\(f = 15\)
Hence, the frequency of the modal class is 15.
In simple words: Substitute all known values into the mode formula and solve the resulting simple linear equation for f.

Exam Tip: Be careful with signs when simplifying the denominator expression \(2f - f_1 - f_2\).

Question 8. The mean of 30 numbers is 18, what will be the new mean, if each observation is increased by 2? (Ans: 20)
Answer:
Let the 30 numbers be \(x_1, x_2, x_3, \dots, x_{30}\).
Given that the mean is 18:
\(\frac{\sum_{i=1}^{30} x_i}{30} = 18\)
\(\sum_{i=1}^{30} x_i = 18 \times 30 = 540\)
If each observation is increased by 2, the new observations become \(x_1 + 2, x_2 + 2, \dots, x_{30} + 2\).
The sum of the new observations is:
\(\sum_{i=1}^{30} (x_i + 2) = \sum_{i=1}^{30} x_i + \sum_{i=1}^{30} 2\)
\(= 540 + (2 \times 30) = 540 + 60 = 600\)
The new mean is:
\(\text{New Mean} = \frac{600}{30} = 20\)
In simple words: When a constant is added to every observation in a dataset, the mean increases by that exact same constant (\(18 + 2 = 20\)).

Exam Tip: Property of mean: If each observation is increased, decreased, multiplied, or divided by a constant \(k\), the mean is affected in the exact same manner.

Question 9. In the graphical representation of a frequency distribution if the distance between mode and mean is k times the distance between median and mean then find the value of k. (Ans: k = 3)
Answer:
Using the empirical formula:
\(\text{Mode} = 3(\text{Median}) - 2(\text{Mean})\)
Subtract Mean from both sides:
\(\text{Mode} - \text{Mean} = 3(\text{Median}) - 2(\text{Mean}) - \text{Mean}\)
\(\text{Mode} - \text{Mean} = 3(\text{Median}) - 3(\text{Mean})\)
\(\text{Mode} - \text{Mean} = 3(\text{Median} - \text{Mean})\)
According to the question, the distance between mode and mean is \(k\) times the distance between median and mean:
\(\text{Mode} - \text{Mean} = k(\text{Median} - \text{Mean})\)
Comparing both equations, we get:
\(k = 3\)
In simple words: Rearrange the standard empirical formula by subtracting the mean from both sides to directly expose the constant factor.

Exam Tip: Remember this standard ratio: the distance from mean to mode is three times the distance from mean to median.

Question 10. Find the mean of 30 numbers given mean of ten of them is 12 and the mean of remaining 20 is 9. (Ans: 10)
Answer:
Total number of observations = 30
Mean of the first 10 numbers = 12
Sum of the first 10 numbers = \(12 \times 10 = 120\)
Mean of the remaining 20 numbers = 9
Sum of the remaining 20 numbers = \(9 \times 20 = 180\)
Total sum of all 30 numbers = \(120 + 180 = 300\)
Combined mean of all 30 numbers =
\(\frac{\text{Total Sum}}{\text{Total Count}} = \frac{300}{30} = 10\)
In simple words: Calculate the sum of each subgroup separately, add them together to get the total sum, and divide by the total count of numbers.

Exam Tip: Combined mean formula: \(\overline{x}_{12} = \frac{n_1\overline{x}_1 + n_2\overline{x}_2}{n_1 + n_2}\). Always find component sums first to avoid calculation errors.