GSEB Class 11 Maths Solutions Chapter 15 Statistics Exercise 15.2

Find the mean and variance for the following data in questions 1 to 5:

 

Question 1. 6, 7, 10, 12, 13, 4, 8, 12
Answer: For the given data, we first need to find the mean and then the variance.
The numbers provided are \( x_i = 6, 7, 10, 12, 13, 4, 8, 12 \).
The total count of these numbers is \( n = 8 \).
The sum of these numbers is \( \Sigma x_i = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72 \).
Mean \( \bar{x} = \frac{\Sigma x_i}{n} = \frac{72}{8} = 9 \).

Now, we calculate the sum of the squared differences from the mean:
\( \Sigma (x_i - \bar{x})^2 = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74 \).
Variance \( \sigma^2 = \frac{\Sigma (x_i - \bar{x})^2}{n} = \frac{74}{8} = 9.25 \).
In simple words: First, sum all the numbers and divide by how many there are to get the average (mean). Then, for each number, subtract the average and square the result. Add up all these squared results, and divide by the total count of numbers to find the variance.

Exam Tip: Remember that variance measures how much the numbers in a data set spread out from their mean. A higher variance means data points are more spread out, while a lower variance means they are closer to the mean.

 

Question 2. First n natural numbers.
Answer: The first n natural numbers are \( 1, 2, 3, \dots, n \).
We know that the sum of the first n natural numbers is \( \Sigma x_i = \frac{n(n+1)}{2} \).
Their mean, \( \bar{x} = \frac{\Sigma x_i}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2} \).
The variance \( \sigma^2 \) is calculated as \( \frac{1}{n} \Sigma x_i^2 - \bar{x}^2 \).
We know that \( \Sigma x_i^2 = \frac{n(n+1)(2n+1)}{6} \).
So, \( \sigma^2 = \frac{1}{n} \left[ \frac{n(n+1)(2n+1)}{6} \right] - \left( \frac{n+1}{2} \right)^2 \)
\( \implies \sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \)
\( \implies \sigma^2 = (n+1) \left[ \frac{2n+1}{6} - \frac{n+1}{4} \right] \)
\( \implies \sigma^2 = (n+1) \left[ \frac{2(2n+1) - 3(n+1)}{12} \right] \)
\( \implies \sigma^2 = (n+1) \left[ \frac{4n+2 - 3n-3}{12} \right] \)
\( \implies \sigma^2 = (n+1) \left[ \frac{n-1}{12} \right] \)
\( \implies \sigma^2 = \frac{n^2-1}{12} \).
In simple words: For a list of numbers starting from 1 up to 'n', the average (mean) is found by adding 1 to 'n' and dividing by 2. The variance, which shows how spread out these numbers are, is calculated by taking 'n' squared, subtracting 1, and then dividing that result by 12.

Exam Tip: Memorizing the formulas for the sum of first n natural numbers and sum of squares of first n natural numbers is crucial for efficiently solving these types of problems.

 

Question 3. First 10 multiples of 3.
Answer: The first 10 multiples of 3 are \( 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 \).
Here, \( n = 10 \). We can use the step deviation method for calculation.
Let the assumed mean \( A = 15 \), and \( h = 3 \).
Then \( y_i = \frac{x_i - A}{h} = \frac{x_i - 15}{3} \).

From the table:
\( \Sigma y_i = 5 \)
\( \Sigma y_i^2 = 85 \)
Mean \( \bar{x} = A + \bar{y} \times h \)
First, find \( \bar{y} = \frac{\Sigma y_i}{n} = \frac{5}{10} = 0.5 \).
So, \( \bar{x} = 15 + 0.5 \times 3 = 15 + 1.5 = 16.5 \).
Variance \( \sigma^2 = \frac{h^2}{n^2} [ n \Sigma y_i^2 - (\Sigma y_i)^2 ] \)
\( \implies \sigma^2 = \frac{3^2}{10^2} [ 10 \times 85 - (5)^2 ] \)
\( \implies \sigma^2 = \frac{9}{100} [ 850 - 25 ] \)
\( \implies \sigma^2 = \frac{9}{100} \times 825 = \frac{7425}{100} = 74.25 \).
Thus, the mean is 16.5 and the variance is 74.25.
In simple words: To find the mean and variance for multiples of 3, we used a shorter method by picking a central number (assumed mean) and a common step size. We created a helper table, calculated the sum of helper values and their squares, and then plugged those into special formulas to get the mean and variance for the actual numbers.

Exam Tip: For data with a common difference (like multiples of a number), the step deviation method significantly simplifies calculations and reduces the chances of errors with large numbers.

 

Question 4.
Answer: For the given frequency distribution, we will find the mean and variance.

From the table:
\( \Sigma f_i = 40 \)
\( \Sigma f_i x_i = 760 \)
Mean \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{760}{40} = 19 \).
Variance \( \sigma^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i} = \frac{1736}{40} = 43.4 \).

**Short-Cut Method:**
We take \( y_i = \frac{x_i - A}{h} \), where \( A = 18 \) and \( h = 2 \). So, \( y_i = \frac{x_i - 18}{2} \).
Now, \( \Sigma f_i y_i = 20 \) and \( \Sigma f_i = 40 \).
Mean \( \bar{x} = A + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h \)
\( \implies \bar{x} = 18 + \frac{20}{40} \times 2 = 18 + \frac{1}{2} \times 2 = 18 + 1 = 19 \).
Variance \( \sigma^2 = \frac{h^2}{(\Sigma f_i)^2} [ \Sigma f_i y_i^2 - n \bar{y}^2 ] \).
Using the formula \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2 ] \)
\( \implies \sigma^2 = \frac{2^2}{40^2} [ 40 \times 444 - (20)^2 ] \)
\( \implies \sigma^2 = \frac{4}{1600} [ 17760 - 400 ] \)
\( \implies \sigma^2 = \frac{1}{400} [ 17360 ] = 43.4 \).
In simple words: We used two methods to find the mean and variance. The first method directly calculated the differences from the mean. The second, shorter method involved transforming the data, using an assumed mean and a step size. Both methods gave the same average (mean) of 19 and a spread (variance) of 43.4.

Exam Tip: Be comfortable with both direct and short-cut methods for calculating mean and variance, as some questions might specifically ask for one or the other. Ensure all calculations are accurate, especially when squaring numbers.

 

Question 5.
Answer: We need to calculate the mean and variance for the provided data.

From the table:
\( \Sigma f_i = 22 \)
\( \Sigma f_i x_i = 2200 \)
Mean \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2200}{22} = 100 \).
Variance \( \sigma^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i} = \frac{640}{22} = 29.09 \).

**Short-Cut Method:**
Let \( y_i = x_i - 98 \), where we assumed \( A = 98 \).
Now, \( \Sigma f_i y_i = 44 \) and \( \Sigma f_i = 22 \).
Mean \( \bar{x} = A + \bar{y} \), where \( \bar{y} = \frac{\Sigma f_i y_i}{\Sigma f_i} = \frac{44}{22} = 2 \).
So, \( \bar{x} = 98 + 2 = 100 \).
Variance \( \sigma^2 = \frac{1}{\Sigma f_i} [ \Sigma f_i y_i^2 - \frac{(\Sigma f_i y_i)^2}{\Sigma f_i} ] \)
\( \implies \sigma^2 = \frac{1}{22} \left[ 728 - \frac{(44)^2}{22} \right] \)
\( \implies \sigma^2 = \frac{1}{22} [ 728 - \frac{1936}{22} ] \)
\( \implies \sigma^2 = \frac{1}{22} [ 728 - 88 ] \)
\( \implies \sigma^2 = \frac{640}{22} = 29.09 \).
In simple words: For this problem, we found the average (mean) to be 100 and the spread (variance) to be 29.09. We used a shorter method by choosing an assumed mean (98) to make the calculations simpler. This involved transforming the data, finding sums of these transformed values, and then using a specific formula to arrive at the final results.

Exam Tip: When using the short-cut method, make sure you correctly identify and use the assumed mean (A) and the common factor (h), if applicable, to avoid errors in the transformed data.

 

Question 6. Find the mean and standard deviation of the following, using short-cut method:

Answer: We will use the short-cut method to find the mean and standard deviation.
Let \( A = 64 \) and \( h = 1 \). So, \( y_i = \frac{x_i - A}{h} = \frac{x_i - 64}{1} \).
From the table:
\( \Sigma f_i = 100 \)
\( \Sigma f_i y_i = 0 \)
\( \Sigma f_i y_i^2 = 286 \)
Mean \( \bar{x} = A + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h \)
\( \implies \bar{x} = 64 + \frac{0}{100} \times 1 = 64 + 0 = 64 \).
Variance \( \sigma^2 = \frac{h^2}{(\Sigma f_i)^2} [ (\Sigma f_i) (\Sigma f_i y_i^2) - (\Sigma f_i y_i)^2 ] \)
\( \implies \sigma^2 = \frac{1^2}{100^2} [ 100 \times 286 - (0)^2 ] \)
\( \implies \sigma^2 = \frac{1}{10000} [ 28600 - 0 ] \)
\( \implies \sigma^2 = \frac{28600}{10000} = 2.86 \).
Standard deviation \( \sigma = \sqrt{\sigma^2} = \sqrt{2.86} = 1.69 \).
In simple words: We used the short-cut method to find the average and spread. We picked 64 as our assumed mean, and a step size of 1. After making a table with transformed values and their frequencies, we calculated the mean as 64 and the variance as 2.86. From the variance, we got the standard deviation, which is 1.69.

Exam Tip: When \( \Sigma f_i y_i \) turns out to be zero in the short-cut method, it significantly simplifies the mean calculation, as the \( \bar{y} \) term becomes zero. This happens when the assumed mean is the actual mean, or very close to it.

 

Find the mean and variance for the following frequency distribution in questions 7 and 8:

 

Question 7.

Answer: We will calculate the mean and variance for the given grouped frequency distribution.
Let \( A = 105 \) and \( h = 30 \). So, \( y_i = \frac{x_i - A}{h} = \frac{x_i - 105}{30} \).
From the table:
\( \Sigma f_i = 30 \)
\( \Sigma f_i y_i = 2 \)
\( \Sigma f_i y_i^2 = 76 \)
Mean \( \bar{x} = A + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h \)
\( \implies \bar{x} = 105 + \frac{2}{30} \times 30 = 105 + 2 = 107 \).
Variance \( \sigma^2 = \frac{h^2}{(\Sigma f_i)^2} [ (\Sigma f_i) (\Sigma f_i y_i^2) - (\Sigma f_i y_i)^2 ] \)
\( \implies \sigma^2 = \frac{30^2}{30^2} [ 30 \times 76 - (2)^2 ] \)
\( \implies \sigma^2 = 1 [ 2280 - 4 ] \)
\( \implies \sigma^2 = 2276 \).
In simple words: We calculated the mean and variance for the provided frequency table. We used the mid-values of each class and applied a short-cut method with an assumed mean of 105 and a class width of 30. After filling in the table and summing up the necessary values, we found the mean to be 107 and the variance to be 2276.

Exam Tip: For grouped data, always accurately find the mid-point (or mid-value) of each class interval, as this represents \( x_i \) in the calculations.

 

Question 8.

Answer: We will calculate the mean and variance for the given frequency distribution.
Let \( A = 25 \) and \( h = 10 \). So, \( y_i = \frac{x_i - A}{h} = \frac{x_i - 25}{10} \).
From the table:
\( \Sigma f_i = 50 \)
\( \Sigma f_i y_i = 10 \)
\( \Sigma f_i y_i^2 = 68 \)
Mean \( \bar{x} = A + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h \)
\( \implies \bar{x} = 25 + \frac{10}{50} \times 10 = 25 + 2 = 27 \).
Variance \( \sigma^2 = \frac{h^2}{(\Sigma f_i)^2} [ (\Sigma f_i) (\Sigma f_i y_i^2) - (\Sigma f_i y_i)^2 ] \)
\( \implies \sigma^2 = \frac{10^2}{50^2} [ 50 \times 68 - (10)^2 ] \)
\( \implies \sigma^2 = \frac{100}{2500} [ 3400 - 100 ] \)
\( \implies \sigma^2 = \frac{1}{25} [ 3300 ] = 132 \).
In simple words: We found the mean and variance for the given data. Using an assumed mean of 25 and a class width of 10, we made a table with transformed values. After calculations, the average (mean) came out to be 27, and the spread (variance) was 132.

Exam Tip: Carefully identify the assumed mean and class width for the step deviation method, as these values are fundamental to correctly transforming the data and simplifying the calculations.

 

Question 9. Find the mean, variance and standard deviation of the following, using short-cut method:

Answer: We will use the short-cut method to find the mean, variance, and standard deviation.
Let \( A = 92.5 \) and \( h = 5 \). So, \( y_i = \frac{x_i - A}{h} = \frac{x_i - 92.5}{5} \).
From the table:
\( \Sigma f_i = 60 \)
\( \Sigma f_i y_i = 6 \)
\( \Sigma f_i y_i^2 = 254 \)
Mean \( \bar{x} = A + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h \)
\( \implies \bar{x} = 92.5 + \frac{6}{60} \times 5 = 92.5 + \frac{1}{10} \times 5 = 92.5 + 0.5 = 93.0 \).
Variance \( \sigma^2 = \frac{h^2}{(\Sigma f_i)^2} [ (\Sigma f_i) (\Sigma f_i y_i^2) - (\Sigma f_i y_i)^2 ] \)
\( \implies \sigma^2 = \frac{5^2}{60^2} [ 60 \times 254 - (6)^2 ] \)
\( \implies \sigma^2 = \frac{25}{3600} [ 15240 - 36 ] \)
\( \implies \sigma^2 = \frac{25}{3600} [ 15204 ] \)
\( \implies \sigma^2 = \frac{15204}{144} = 105.58 \).
Standard deviation \( \sigma = \sqrt{\sigma^2} = \sqrt{105.58} = 10.27 \).
In simple words: Using the short-cut method, we found the average height (mean) to be 93.0 cm, the variance (spread) to be 105.58, and the standard deviation to be 10.27. We set the assumed mean at 92.5 and a class width of 5 to simplify the calculations, which helped in efficiently processing the data.

Exam Tip: Always double-check your calculations, especially with squares and large sums, as a small arithmetic error early on can propagate and lead to an incorrect final answer.

 

Question 10. The diameter of circles (in mm) drawn in a design are given below: Calculate the mean diameter and standard deviation of the circles.

Answer: Firstly, the data needs to be made continuous by adjusting the classes. The adjusted classes will be: \( 32.5 - 36.5, 36.5 - 40.5, 40.5 - 44.5, 44.5 - 48.5, 48.5 - 52.5 \).
We will use the short-cut method to find the mean diameter and standard deviation.
Let \( A = 42.5 \) and \( h = 4 \). So, \( y_i = \frac{x_i - A}{h} = \frac{x_i - 42.5}{4} \).
From the table:
\( \Sigma f_i = 100 \)
\( \Sigma f_i y_i = 25 \)
\( \Sigma f_i y_i^2 = 199 \)
Mean \( \bar{x} = A + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h \)
\( \implies \bar{x} = 42.5 + \frac{25}{100} \times 4 = 42.5 + 1 = 43.5 \).
Variance \( \sigma^2 = \frac{h^2}{(\Sigma f_i)^2} [ (\Sigma f_i) (\Sigma f_i y_i^2) - (\Sigma f_i y_i)^2 ] \)
\( \implies \sigma^2 = \frac{4^2}{100^2} [ 100 \times 199 - (25)^2 ] \)
\( \implies \sigma^2 = \frac{16}{10000} [ 19900 - 625 ] \)
\( \implies \sigma^2 = \frac{16}{10000} [ 19275 ] \)
\( \implies \sigma^2 = \frac{308400}{10000} = 30.84 \).
Standard deviation \( \sigma = \sqrt{\sigma^2} = \sqrt{30.84} = 5.55 \).
In simple words: First, we made the class intervals continuous. Then, using the short-cut method, we picked an assumed mean of 42.5 and a class width of 4. After creating a table with transformed values and performing the necessary calculations, we found the average diameter (mean) to be 43.5 mm and the standard deviation to be 5.55 mm, which indicates the spread of the diameters.

Exam Tip: Always make sure to convert discontinuous class intervals into continuous ones before attempting to calculate mean or variance for grouped frequency distributions.