Samacheer Kalvi Class 10 Maths Solutions Chapter 8 Statistics and Probability Exercise 8.1

Question 1. Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) For the data set: 63, 89, 98, 125, 79, 108, 117, 68
The largest value (L) is 125.
The smallest value (S) is 63.
Now, we calculate the range and coefficient of range.
Range \( = L - S = 125 - 63 = 62 \)
Coefficient of range \( = \frac{L - S}{L + S} = \frac{125 - 63}{125 + 63} = \frac{62}{188} = 0.33 \)
(ii) For the data set: 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
The largest value (L) is 61.4.
The smallest value (S) is 13.6.
Range \( = L - S = 61.4 - 13.6 = 47.8 \)
Coefficient of range \( = \frac{L - S}{L + S} = \frac{61.4 - 13.6}{61.4 + 13.6} = \frac{47.8}{75} = 0.64 \)
In simple words: The range tells us how spread out the data is by finding the difference between the biggest and smallest numbers. The coefficient of range gives us a relative measure of this spread.

🎯 Exam Tip: Remember to always identify the largest and smallest values correctly from the given data, as a small error can lead to wrong calculations for both range and coefficient of range.

Question 2. If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Answer:
We are given the range and the smallest value of a data set.
Range \( = 36.8 \)
Smallest value (S) \( = 13.4 \)
We know that the formula for range is: Range \( = L - S \), where L is the largest value and S is the smallest value.
\( 36.8 = L - 13.4 \)
To find L, we add S to the range:
\( L = \text{Range} + S \)
\( L = 36.8 + 13.4 \)
\( L = 50.2 \)
So, the largest value is 50.2.
In simple words: If you know how wide a group of numbers is (the range) and where it starts (the smallest number), you can find where it ends (the largest number) by just adding them together.

🎯 Exam Tip: Always remember the basic formula for range, Range = Largest Value - Smallest Value. This helps in easily finding any unknown variable when the other two are given.

Question 3. Calculate the range of the following data.

Answer:
For the given data, we need to find the range. The range is the difference between the largest and smallest values in the data set.
The smallest income group is 400-450, so the smallest value (S) is 400.
The largest income group is 600-650, so the largest value (L) is 650.
The formula for range is: Range \( = L - S \)
Substituting the values, we get:
Range \( = 650 - 400 \)
Range \( = 250 \)
So, the range of the data is 250.
In simple words: To find the range of grouped data, take the highest value from the last group and subtract the lowest value from the first group. This shows the total spread of all incomes.

🎯 Exam Tip: When calculating the range for grouped data, always use the lower limit of the first class interval as the smallest value and the upper limit of the last class interval as the largest value.

Question 4. A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer:
First, we need to find out how many pages each student still needs to complete. Each student has to complete 60 pages.
The pages completed by eight students are: 32, 35, 37, 30, 33, 36, 35, 37.
The remaining number of pages to be completed for each student are:
\( 60 - 32 = 28 \)
\( 60 - 35 = 25 \)
\( 60 - 37 = 23 \)
\( 60 - 30 = 30 \)
\( 60 - 33 = 27 \)
\( 60 - 36 = 24 \)
\( 60 - 35 = 25 \)
\( 60 - 37 = 23 \)
So, the pages yet to be completed are: 28, 25, 23, 30, 27, 24, 25, and 23.
Arranging these values in ascending order: 23, 23, 24, 25, 25, 27, 28, 30.
Here, the number of observations (n) = 8.
Now, we create a table to calculate \(\Sigma x_i\) and \(\Sigma x_i^2\):

We have n = 8, \( \Sigma x_i = 205 \), and \( \Sigma x_i^2 = 5297 \).
The formula for standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma x_i^2}{n} - \left(\frac{\Sigma x_i}{n}\right)^2} \)
Substitute the values into the formula:
\( \sigma = \sqrt{\frac{5297}{8} - \left(\frac{205}{8}\right)^2} \)
\( \sigma = \sqrt{662.125 - (25.625)^2} \)
\( \sigma = \sqrt{662.125 - 656.640625} \)
\( \sigma = \sqrt{5.484375} \)
\( \sigma \approx 2.3418 \)
Rounding to two decimal places, the standard deviation \( (\sigma) = 2.34 \).
In simple words: First, find how many pages each student still needs to complete. Then, use these new numbers to calculate the standard deviation, which shows how much these "remaining pages" numbers vary from the average.

🎯 Exam Tip: Be careful with calculations involving decimals and squares, especially in the standard deviation formula. It's often helpful to keep a few extra decimal places during intermediate steps to ensure accuracy in the final answer.

Question 5. Find the variance and standard deviation of the wages of 9 workers given below: Rs 310, Rs 290, Rs 320, Rs 280, Rs 300, Rs 290, Rs 320, Rs 310, Rs 280.
Answer:
The wages of 9 workers are: Rs 310, Rs 290, Rs 320, Rs 280, Rs 300, Rs 290, Rs 320, Rs 310, Rs 280.
First, arrange the wages in ascending order:
280, 280, 290, 290, 300, 310, 310, 320, 320.
Here, the number of observations (n) = 9.
We will use the assumed mean method. Let the assumed mean (A) = 300.
Now, we create a table to calculate \(d_i = x_i - A\) and \(d_i^2\):

Here, n = 9, \( \Sigma d_i = 0 \), and \( \Sigma d_i^2 = 2000 \).
The formula for variance \( (\sigma^2) \) using the assumed mean method is:
\( \sigma^2 = \frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2 \)
Substitute the values:
\( \sigma^2 = \frac{2000}{9} - \left(\frac{0}{9}\right)^2 \)
\( \sigma^2 = \frac{2000}{9} - 0 \)
\( \sigma^2 = 222.222... \)
Rounding to three decimal places, Variance \( (\sigma^2) = 222.222 \).
Now, to find the standard deviation \( (\sigma) \), we take the square root of the variance:
\( \sigma = \sqrt{\text{Variance}} = \sqrt{222.222} \)
\( \sigma = 14.90709... \)
Rounding to two decimal places, Standard deviation \( (\sigma) = 14.91 \).
In simple words: We find how much each worker's wage differs from an average wage we picked. Then we use these differences to calculate variance, which shows how spread out the wages are. The square root of variance gives us standard deviation, which is another way to measure how wages vary.

🎯 Exam Tip: The assumed mean method simplifies calculations, especially when the data values are large. Choose an assumed mean close to the middle of the data set to keep \(d_i\) values small.

Question 6. A wall clock strikes the bell once at 1 o'clock, 2 times at 2 o'clock, 3 times at 3 o'clock and so on. How many times will it strike in a particular day? Find the Standard deviation of the number of strikes the bell make a day.
Answer:
A wall clock strikes once at 1, twice at 2, and so on. This pattern repeats for every 12 hours.
In 12 hours, the total number of strikes is: \( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 \).
The sum of the first n natural numbers is \( \frac{n(n+1)}{2} \).
For 12 hours: \( \frac{12(12+1)}{2} = \frac{12 \times 13}{2} = 6 \times 13 = 78 \) strikes.
A day has 24 hours, so the clock will strike for two 12-hour cycles.
Total strikes in a day \( = 78 \times 2 = 156 \) strikes.

Now, let's find the standard deviation of the number of strikes in a day. The strikes occur as 1, 2, ..., 12 in the first 12 hours and then repeat for the next 12 hours. So, the data for strikes in a day (24 hours) is: 1, 2, ..., 12, then again 1, 2, ..., 12. Alternatively, for 24 hours, the question means the counts are 2, 4, 6, 8, ..., 24, as the clock strikes twice for each hour value in a day.
Let's consider the number of strikes as the number of events (2 times at 2 o'clock means it strikes 2 bells, not just 1). In a day, the clock repeats its 12-hour cycle. So the strike counts are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and then again 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
However, the standard approach for this kind of question usually considers the data points as a list of strike counts per hour over 24 hours (e.g., 1 strike at 1 AM, 2 at 2 AM, ..., 12 at 12 PM, then 1 at 1 PM, etc.). The provided solution uses a sequence of even numbers 2, 4, 6... for 24 hours, implying the question asks for the standard deviation of strikes on an hourly basis throughout the day where 1-o'clock means 2 strikes, 2-o'clock means 4 strikes, and so on, over 12 unique hours. Let's follow the provided solution's interpretation for calculating standard deviation, which lists the strike values as 2, 4, 6, ..., 24 for the 12 hours of the clock in a 24-hour cycle. This means at 1 it strikes 2 times (1x2), at 2 it strikes 4 times (2x2) up to 12 it strikes 24 times (12x2).
The strike values (\(x_i\)) are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Number of observations (n) = 12.
Let the assumed mean (A) = 14.
Now, we create a table to calculate \(d_i = x_i - A\) and \(d_i^2\):

Here, n = 12, \( \Sigma d_i = -12 \), and \( \Sigma d_i^2 = 584 \).
The formula for standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \)
Substitute the values:
\( \sigma = \sqrt{\frac{584}{12} - \left(\frac{-12}{12}\right)^2} \)
\( \sigma = \sqrt{48.666... - (-1)^2} \)
\( \sigma = \sqrt{48.666... - 1} \)
\( \sigma = \sqrt{47.666...} \)
\( \sigma \approx 6.904 \)
Rounding to one decimal place, Standard deviation \( (\sigma) = 6.9 \).

Aliter (Alternative method for standard deviation):
The strike numbers in a day are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
These are even numbers, which can be written as \( 2 \times (1, 2, 3, ..., 12) \).
So, we have \( 2 \times n \) for \( n = 1, 2, ..., 12 \).
The standard deviation for the first 'n' natural numbers is \( \sqrt{\frac{n^2 - 1}{12}} \).
Here, the data is \( 2 \times (1, 2, ..., 12) \), which means the standard deviation of this data will be \( 2 \times \) the standard deviation of \( (1, 2, ..., 12) \).
For \( n = 12 \), the standard deviation of natural numbers is:
\( \text{S.D.} = \sqrt{\frac{12^2 - 1}{12}} = \sqrt{\frac{144 - 1}{12}} = \sqrt{\frac{143}{12}} = \sqrt{11.9166...} \approx 3.4578 \)
Therefore, the standard deviation for the given strike values is \( 2 \times 3.4578 \approx 6.9156 \).
Rounding to one decimal place, \( 6.9 \).
In simple words: First, calculate the total number of times the clock strikes in a full day by adding up strikes for each hour and doubling for 24 hours. Then, to find the standard deviation, we treat the hourly strike counts as data points and see how much they vary from their average. We can use an assumed mean to make calculations easier.

🎯 Exam Tip: When dealing with data that is a multiple of natural numbers (like 2, 4, 6, ...), you can sometimes simplify the standard deviation calculation by finding the SD of the natural numbers and then multiplying by the common factor.

Question 7. Find the standard deviation of the first 21 natural numbers.
Answer:
We need to find the standard deviation of the first 21 natural numbers.
Here, the number of natural numbers (n) = 21.
The formula for the standard deviation \( (\sigma) \) of the first 'n' natural numbers is:
\( \sigma = \sqrt{\frac{n^2 - 1}{12}} \)
Substitute \( n = 21 \) into the formula:
\( \sigma = \sqrt{\frac{21^2 - 1}{12}} \)
\( \sigma = \sqrt{\frac{441 - 1}{12}} \)
\( \sigma = \sqrt{\frac{440}{12}} \)
\( \sigma = \sqrt{36.666...} \)
\( \sigma \approx 6.0556 \)
Rounding to two decimal places, the standard deviation \( (\sigma) = 6.06 \).
In simple words: To find how spread out the first few counting numbers are, we use a special formula that quickly gives us the standard deviation based on how many numbers there are.

🎯 Exam Tip: Memorizing the formula for the standard deviation of the first 'n' natural numbers, \( \sigma = \sqrt{\frac{n^2 - 1}{12}} \), can save significant time on such questions in an exam.

Question 8. If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Answer:
We are given that the standard deviation of a data set is 4.5.
The problem states that each value in the data set is decreased by 5.
When a constant value is added to or subtracted from each observation in a data set, the standard deviation remains unchanged. This is because standard deviation measures the spread or dispersion of data points around the mean. Shifting all data points by the same amount does not change their relative distances from each other, nor does it change their distances from the new mean (which also shifts by the same amount).
Therefore, if each value of the data is decreased by 5, the new standard deviation will still be 4.5.
In simple words: Standard deviation tells us how spread out numbers are. If we move all the numbers by the same amount (like subtracting 5 from each), they stay just as spread out, so the standard deviation does not change.

🎯 Exam Tip: Remember this key property: adding or subtracting a constant to every data point does not affect the standard deviation or variance, only the mean. However, multiplying or dividing by a constant *does* affect it.

Question 9. If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer:
We are given the standard deviation of a data set \( (\sigma) = 3.6 \).
Each value of the data is divided by 3.
When each value in a data set is divided by a constant 'k', the new standard deviation becomes the original standard deviation divided by 'k'.
So, the new standard deviation \( (\sigma_{\text{new}}) = \frac{\text{Original Standard Deviation}}{3} \)
\( \sigma_{\text{new}} = \frac{3.6}{3} = 1.2 \)
Now, we need to find the new variance.
The variance \( (\sigma^2) \) is the square of the standard deviation.
New Variance \( = (\text{New Standard Deviation})^2 \)
New Variance \( = (1.2)^2 = 1.44 \)
So, the new standard deviation is 1.2, and the new variance is 1.44.
In simple words: If you divide every number in a list by a number (like 3), the spread of those numbers (standard deviation) also gets divided by that same number. The variance, which is the spread squared, will get divided by that number squared.

🎯 Exam Tip: Be careful to distinguish between the effect of addition/subtraction and multiplication/division on standard deviation and variance. Only multiplication/division changes the spread, while addition/subtraction only shifts the data.

Question 10. The rainfall recorded in various places of five districts in a week are given below. Find its standard deviation.

Answer:
We are given the rainfall data and the number of places. This is grouped data.
Let \(x_i\) be the rainfall (in mm) and \(f_i\) be the number of places.
We will use the assumed mean method to calculate the standard deviation.
Let the assumed mean (A) = 60.
Now, we construct a table to calculate \(d_i = x_i - A\), \(f_id_i\), and \(f_id_i^2\):

Here, \( N = 40 \), \( \Sigma f_id_i = -160 \), and \( \Sigma f_id_i^2 = 3050 \).
The formula for standard deviation \( (\sigma) \) for grouped data using the assumed mean method is:
\( \sigma = \sqrt{\frac{\Sigma f_id_i^2}{N} - \left(\frac{\Sigma f_id_i}{N}\right)^2} \)
Substitute the values into the formula:
\( \sigma = \sqrt{\frac{3050}{40} - \left(\frac{-160}{40}\right)^2} \)
\( \sigma = \sqrt{76.25 - (-4)^2} \)
\( \sigma = \sqrt{76.25 - 16} \)
\( \sigma = \sqrt{60.25} \)
\( \sigma = 7.762 \)
Rounding to two decimal places, the standard deviation \( (\sigma) = 7.76 \).
In simple words: To find how much rainfall varies across different places, we use a method where we pick an average value (assumed mean), find the differences from this average for each rainfall amount, and then use these differences to calculate the standard deviation.

🎯 Exam Tip: For grouped data, remember to multiply the deviations and their squares by the frequency (\(f_i\)) of each class. The total frequency N is used in the denominator, not just the count of unique \(x_i\) values.

Question 11. In a study about viral fever, the number of people affected in a town were noted as:

Find its standard deviation.
Answer:
We are given the age groups and the number of people affected. This is continuous grouped data.
First, we find the mid-value (\(x_i\)) for each age group. Then, we use the assumed mean method.
Let the assumed mean (A) = 35 (the mid-value of the 30-40 age group).
Now, we construct a detailed table to calculate \(d_i = x_i - A\), \(f_id_i\), and \(f_id_i^2\):

Here, \( N = \Sigma f_i = 65 \), \( \Sigma f_id_i = 30 \), and \( \Sigma f_id_i^2 = 13900 \).
The formula for standard deviation \( (\sigma) \) for grouped data using the assumed mean method is:
\( \sigma = \sqrt{\frac{\Sigma f_id_i^2}{N} - \left(\frac{\Sigma f_id_i}{N}\right)^2} \)
Substitute the values into the formula:
\( \sigma = \sqrt{\frac{13900}{65} - \left(\frac{30}{65}\right)^2} \)
\( \sigma = \sqrt{213.846... - (0.4615...)^2} \)
\( \sigma = \sqrt{213.846... - 0.213...} \)
\( \sigma = \sqrt{213.633...} \)
\( \sigma = 14.616... \)
Rounding to two decimal places, the standard deviation \( (\sigma) = 14.62 \).
In simple words: We find the middle point for each age group. Then, we choose an assumed average age and calculate how much each group's middle age differs from it. Using these differences and the number of affected people, we figure out the standard deviation, which shows how much the ages of affected people vary.

🎯 Exam Tip: For continuous grouped data, correctly calculating the mid-value (\(x_i\)) for each class interval is crucial. If the intervals are inclusive (e.g., 0-10, 11-20), adjust them to exclusive (0-10, 10-20) or be consistent in mid-point calculation.

Question 12. The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Answer:
We are given the diameter data and the number of plates. This is continuous grouped data.
First, we find the mid-value (\(x_i\)) for each diameter group. Then, we use the assumed mean method.
Let the assumed mean (A) = 34.5 (the mid-value of the 33-36 diameter group).
Now, we construct a detailed table to calculate \(d_i = x_i - A\), \(f_id_i\), and \(f_id_i^2\):

Here, \( N = \Sigma f_i = 84 \), \( \Sigma f_id_i = -316 \), and \( \Sigma f_id_i^2 = 4208 \).
The formula for standard deviation \( (\sigma) \) for grouped data using the assumed mean method is:
\( \sigma = \sqrt{\frac{\Sigma f_id_i^2}{N} - \left(\frac{\Sigma f_id_i}{N}\right)^2} \)
Substitute the values into the formula:
\( \sigma = \sqrt{\frac{4208}{84} - \left(\frac{-316}{84}\right)^2} \)
\( \sigma = \sqrt{50.0952... - (-3.7619...)^2} \)
\( \sigma = \sqrt{50.0952... - 14.1519...} \)
\( \sigma = \sqrt{35.9433...} \)
\( \sigma = 5.9952... \)
Rounding to the nearest whole number (as per source), the standard deviation \( (\sigma) = 6 \).
In simple words: To find how much the diameters of plates vary, we calculate the middle value for each diameter range. Then, we use an assumed average diameter to find how far each group's middle value is from it. These differences help us calculate the standard deviation, showing the spread of plate sizes.

🎯 Exam Tip: When dealing with inclusive class intervals like 21-24, remember that the mid-value is calculated as (lower limit + upper limit) / 2. Also, choose an assumed mean that is a mid-value to simplify calculations of \(d_i\).

Question 13. The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Answer:
We are given the time taken and the number of students. This is continuous grouped data.
First, we find the mid-value (\(x_i\)) for each time interval. Then, we use the assumed mean method.
Let the assumed mean (A) = 11 (the mid-value of the 10.5-11.5 time group).
Now, we construct a detailed table to calculate \(d_i = x_i - A\), \(f_id_i\), and \(f_id_i^2\):

Here, \( N = \Sigma f_i = 50 \), \( \Sigma f_id_i = 8 \), and \( \Sigma f_id_i^2 = 78 \).
The formula for standard deviation \( (\sigma) \) for grouped data using the assumed mean method is:
\( \sigma = \sqrt{\frac{\Sigma f_id_i^2}{N} - \left(\frac{\Sigma f_id_i}{N}\right)^2} \)
Substitute the values into the formula:
\( \sigma = \sqrt{\frac{78}{50} - \left(\frac{8}{50}\right)^2} \)
\( \sigma = \sqrt{1.56 - (0.16)^2} \)
\( \sigma = \sqrt{1.56 - 0.0256} \)
\( \sigma = \sqrt{1.5344} \)
\( \sigma = 1.2386... \)
Rounding to two decimal places, the standard deviation \( (\sigma) = 1.24 \).
In simple words: For grouped data like race times, we calculate the middle time for each group. We then use an assumed average time to find how much each group's middle time differs. These differences, along with how many students are in each group, help us find the standard deviation, which shows how varied the race times are.

🎯 Exam Tip: Pay close attention to inclusive vs. exclusive class intervals when calculating mid-values. In exclusive intervals (like 9.5-10.5), the mid-value is straightforward. If it were 9-10 and 11-12, the mid-values would be 9.5 and 11.5 respectively.

Question 14. For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer:
We are given:
Number of candidates (n) = 100
Incorrect Mean \( (\bar{x}) = 60 \)
Incorrect Standard Deviation \( (\sigma) = 15 \)
Wrongly entered scores: 40 and 27
Correct scores: 45 and 72

First, let's find the incorrect sum of marks \( (\Sigma x) \):
Mean \( = \frac{\Sigma x}{n} \)
\( 60 = \frac{\Sigma x}{100} \)
\( \Sigma x = 60 \times 100 = 6000 \)

Now, let's find the correct sum of marks \( (\Sigma x_{\text{correct}}) \):
\( \Sigma x_{\text{correct}} = \Sigma x_{\text{incorrect}} - (\text{wrong entries}) + (\text{correct entries}) \)
\( \Sigma x_{\text{correct}} = 6000 - (40 + 27) + (45 + 72) \)
\( \Sigma x_{\text{correct}} = 6000 - 67 + 117 \)
\( \Sigma x_{\text{correct}} = 6000 + 50 = 6050 \)

Calculate the correct mean:
Correct Mean \( (\bar{x}_{\text{correct}}) = \frac{\Sigma x_{\text{correct}}}{n} = \frac{6050}{100} = 60.5 \)

Next, let's find the incorrect sum of squares \( (\Sigma x^2) \). We use the formula for standard deviation:
\( \sigma = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} \)
\( 15 = \sqrt{\frac{\Sigma x^2}{100} - \left(\frac{6000}{100}\right)^2} \)
\( 15 = \sqrt{\frac{\Sigma x^2}{100} - (60)^2} \)
Square both sides:
\( 15^2 = \frac{\Sigma x^2}{100} - 3600 \)
\( 225 = \frac{\Sigma x^2}{100} - 3600 \)
\( \frac{\Sigma x^2}{100} = 225 + 3600 = 3825 \)
\( \Sigma x^2 = 3825 \times 100 = 382500 \)

Now, let's find the correct sum of squares \( (\Sigma x^2_{\text{correct}}) \):
\( \Sigma x^2_{\text{correct}} = \Sigma x^2_{\text{incorrect}} - (\text{wrong entries})^2 + (\text{correct entries})^2 \)
\( \Sigma x^2_{\text{correct}} = 382500 - (40^2 + 27^2) + (45^2 + 72^2) \)
\( \Sigma x^2_{\text{correct}} = 382500 - (1600 + 729) + (2025 + 5184) \)
\( \Sigma x^2_{\text{correct}} = 382500 - 2329 + 7209 \)
\( \Sigma x^2_{\text{correct}} = 380171 + 7209 = 387380 \)

Finally, calculate the correct standard deviation \( (\sigma_{\text{correct}}) \):
\( \sigma_{\text{correct}} = \sqrt{\frac{\Sigma x^2_{\text{correct}}}{n} - \left(\frac{\Sigma x_{\text{correct}}}{n}\right)^2} \)
\( \sigma_{\text{correct}} = \sqrt{\frac{387380}{100} - \left(\frac{6050}{100}\right)^2} \)
\( \sigma_{\text{correct}} = \sqrt{3873.8 - (60.5)^2} \)
\( \sigma_{\text{correct}} = \sqrt{3873.8 - 3660.25} \)
\( \sigma_{\text{correct}} = \sqrt{213.55} \)
\( \sigma_{\text{correct}} = 14.613... \)
Rounding to two decimal places, the correct standard deviation \( (\sigma_{\text{correct}}) = 14.61 \).
In simple words: First, we use the wrong mean to find the total sum of marks. Then, we fix this sum by removing the incorrect marks and adding the correct ones, which gives us the correct mean. Next, we use the wrong standard deviation to find the total sum of squared marks. We fix this sum by removing the squares of the wrong marks and adding the squares of the correct marks. Finally, we use this corrected sum of squares and the correct mean to find the accurate standard deviation.

🎯 Exam Tip: When correcting mean and standard deviation for errors, remember to adjust both the sum of observations (\(\Sigma x\)) and the sum of squares of observations (\(\Sigma x^2\)). Don't forget to square the individual values when adjusting \(\Sigma x^2\).

Question 15. The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer:
We are given:
Number of observations (n) = 7
Mean \( (\bar{x}) = 8 \)
Variance \( (\sigma^2) = 16 \)
Five observations: 2, 4, 10, 12, 14
Let the remaining two observations be 'a' and 'b'.
So, the seven observations are: 2, 4, 10, 12, 14, a, b.

First, let's use the mean formula:
Mean \( = \frac{\Sigma x_i}{n} \)
\( 8 = \frac{2 + 4 + 10 + 12 + 14 + a + b}{7} \)
\( 8 = \frac{42 + a + b}{7} \)
Multiply both sides by 7:
\( 56 = 42 + a + b \)
\( a + b = 56 - 42 \)
\( a + b = 14 \) ... (Equation 1)

Now, let's use the variance formula:
Variance \( = \frac{\Sigma x_i^2}{n} - \left(\frac{\Sigma x_i}{n}\right)^2 \)
We know \( \frac{\Sigma x_i}{n} \) is the mean, which is 8. So, \( \left(\frac{\Sigma x_i}{n}\right)^2 = (8)^2 = 64 \).
\( 16 = \frac{\Sigma x_i^2}{7} - (8)^2 \)
\( 16 = \frac{\Sigma x_i^2}{7} - 64 \)
\( 16 + 64 = \frac{\Sigma x_i^2}{7} \)
\( 80 = \frac{\Sigma x_i^2}{7} \)
\( \Sigma x_i^2 = 80 \times 7 = 560 \)

Now, let's calculate \( \Sigma x_i^2 \) from the individual observations:
\( \Sigma x_i^2 = 2^2 + 4^2 + 10^2 + 12^2 + 14^2 + a^2 + b^2 \)
\( \Sigma x_i^2 = 4 + 16 + 100 + 144 + 196 + a^2 + b^2 \)
\( \Sigma x_i^2 = 460 + a^2 + b^2 \)
Equating this to the value we found for \( \Sigma x_i^2 \):
\( 460 + a^2 + b^2 = 560 \)
\( a^2 + b^2 = 560 - 460 \)
\( a^2 + b^2 = 100 \) ... (Equation 2)

We have a system of two equations:
1) \( a + b = 14 \)
2) \( a^2 + b^2 = 100 \)
From Equation 1, we can write \( b = 14 - a \).
Substitute this into Equation 2:
\( a^2 + (14 - a)^2 = 100 \)
\( a^2 + (14^2 - 2 \times 14 \times a + a^2) = 100 \)
\( a^2 + 196 - 28a + a^2 = 100 \)
\( 2a^2 - 28a + 196 - 100 = 0 \)
\( 2a^2 - 28a + 96 = 0 \)
Divide the entire equation by 2:
\( a^2 - 14a + 48 = 0 \)
This is a quadratic equation. We can factorize it:
We need two numbers that multiply to 48 and add to -14. These numbers are -6 and -8.
\( (a - 6)(a - 8) = 0 \)
So, \( a = 6 \) or \( a = 8 \).

If \( a = 6 \), then from \( a + b = 14 \):
\( 6 + b = 14 \implies b = 8 \)
If \( a = 8 \), then from \( a + b = 14 \):
\( 8 + b = 14 \implies b = 6 \)
Thus, the two remaining observations are 6 and 8.
In simple words: We use the given mean to find the total sum of all seven numbers. Then, we use the variance to find the total sum of the squares of all seven numbers. We set up two equations using these totals and the two unknown numbers. By solving these equations, we find the values of the two missing numbers.

🎯 Exam Tip: When solving for two unknown values using mean and variance, remember that you'll always get two equations: one linear from the mean (sum of values) and one quadratic from the variance (sum of squares). Carefully solve the quadratic equation to find the possible values.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.1

Question 1. Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) Here, for the data set \( \{63, 89, 98, 125, 79, 108, 117, 68\} \):
Largest value \( (L) = 125 \)
Smallest value \( (S) = 63 \)
Range \( = L - S = 125 - 63 = 62 \)
Coefficient of range \( = \frac{L - S}{L + S} = \frac{125 - 63}{125 + 63} = \frac{62}{188} \approx 0.33 \)
(ii) Here, for the data set \( \{43.5, 13.6, 18.9, 38.4, 61.4, 29.8\} \):
Largest value \( (L) = 61.4 \)
Smallest value \( (S) = 13.6 \)
Range \( = L - S = 61.4 - 13.6 = 47.8 \)
Coefficient of range \( = \frac{L - S}{L + S} = \frac{61.4 - 13.6}{61.4 + 13.6} = \frac{47.8}{75} \approx 0.64 \)
In simple words: The range tells us how spread out the data is, found by subtracting the smallest number from the largest. The coefficient of range helps compare the spread of different data sets, calculated by dividing the range by the sum of the largest and smallest numbers.

🎯 Exam Tip: Always remember to sort the data first (or at least identify the largest and smallest values correctly) before calculating the range or coefficient of range.

Question 2. If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Answer: We know that Range \( = \text{Largest Value} - \text{Smallest Value} \).
Given Range \( = 36.8 \)
Given Smallest Value \( (S) = 13.4 \)
So, \( 36.8 = \text{Largest Value} - 13.4 \)
To find the largest value, we add the range and the smallest value.
Largest Value \( = 36.8 + 13.4 = 50.2 \)
In simple words: If you know how much a set of numbers spreads out (the range) and what the smallest number is, you can find the largest number by adding these two values together. This simple calculation gives us the upper limit of the data spread.

🎯 Exam Tip: Remember the basic formula for range and rearrange it algebraically to find any missing component, such as the largest or smallest value.

Question 3. Calculate the range of the following data.

Answer: To find the range for grouped data, we use the class limits.
The smallest value \( (S) \) is the lower limit of the first class interval \( = 400 \).
The largest value \( (L) \) is the upper limit of the last class interval \( = 650 \).
Range \( = L - S = 650 - 400 = 250 \).
In simple words: For data grouped into classes, the range is simply the highest value possible in the last group minus the lowest value possible in the first group. It gives the total spread of all the data.

🎯 Exam Tip: When dealing with grouped frequency data, the range is typically found by taking the upper limit of the highest class interval and subtracting the lower limit of the lowest class interval.

Question 4. A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer: First, we find the number of pages each student still needs to complete.
Total pages to complete \( = 60 \).
Completed pages are \( \{32, 35, 37, 30, 33, 36, 35, 37\} \).
Remaining pages to be completed are:
\( 60 - 32 = 28 \)
\( 60 - 35 = 25 \)
\( 60 - 37 = 23 \)
\( 60 - 30 = 30 \)
\( 60 - 33 = 27 \)
\( 60 - 36 = 24 \)
\( 60 - 35 = 25 \)
\( 60 - 37 = 23 \)
So, the pages to be completed are: \( \{28, 25, 23, 30, 27, 24, 25, 23\} \).
Arranging these in ascending order: \( \{23, 23, 24, 25, 25, 27, 28, 30\} \).
We have \( n = 8 \) (number of students).
Now we calculate standard deviation using the direct method for individual data:

Answer: (Continued)
Sum of \( x_i \): \( \Sigma x_i = 23 + 23 + 24 + 25 + 25 + 27 + 28 + 30 = 205 \)
Sum of \( x_i^2 \): \( \Sigma x_i^2 = 529 + 529 + 576 + 625 + 625 + 729 + 784 + 900 = 5297 \)
Here, \( n = 8 \), \( \Sigma x_i = 205 \), \( \Sigma x_i^2 = 5297 \).
The formula for Standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma x_i^2}{n} - \left(\frac{\Sigma x_i}{n}\right)^2} \)
\( \implies \sigma = \sqrt{\frac{5297}{8} - \left(\frac{205}{8}\right)^2} \)
\( \implies \sigma = \sqrt{662.125 - \left(\frac{42025}{64}\right)} \)
\( \implies \sigma = \sqrt{662.125 - 656.640625} \)
\( \implies \sigma = \sqrt{5.484375} \)
\( \implies \sigma \approx 2.34 \)
So, the standard deviation of the pages yet to be completed is approximately \( 2.34 \).
In simple words: First, figure out how many pages each student still has to do. Then, use a special formula that looks at each number and its square to calculate how much these remaining pages vary from the average. This helps us understand how consistent the students are in their incompleteness.

🎯 Exam Tip: When subtracting from a total (like 60 pages), be careful to calculate each individual difference accurately before proceeding to statistical calculations.

Question 5. Find the variance and standard deviation of the wages of 9 workers given below: Rs 310, Rs 290, Rs 320, Rs 280, Rs 300, Rs 290, Rs 320, Rs 310, Rs 280.
Answer: First, arrange the wages in ascending order:
Rs \( \{280, 280, 290, 290, 300, 310, 310, 320, 320\} \).
We have \( n = 9 \) (number of workers).
Let's use the assumed mean method. Let the assumed mean \( (A) = 300 \).

Answer: (Continued)
Sum of deviations: \( \Sigma d_i = (-20) \times 2 + (-10) \times 2 + 0 \times 1 + 10 \times 2 + 20 \times 2 = -40 - 20 + 0 + 20 + 40 = 0 \)
Sum of squared deviations: \( \Sigma d_i^2 = 400 \times 2 + 100 \times 2 + 0 + 100 \times 2 + 400 \times 2 = 800 + 200 + 0 + 200 + 800 = 2000 \)
Here \( n = 9 \), \( \Sigma d_i = 0 \), \( \Sigma d_i^2 = 2000 \).
The formula for Variance is:
Variance \( = \frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2 \)
\( \implies \) Variance \( = \frac{2000}{9} - \left(\frac{0}{9}\right)^2 \)
\( \implies \) Variance \( = 222.222 - 0 \)
\( \implies \) Variance \( = 222.222 \)
The formula for Standard deviation \( (\sigma) \) is:
Standard deviation \( = \sqrt{\text{Variance}} \)
\( \implies \) Standard deviation \( = \sqrt{222.222} \)
\( \implies \) Standard deviation \( \approx 14.907 \approx 14.91 \)
So, the variance is \( 222.222 \) and the standard deviation is \( 14.91 \). These measures help us understand the spread of wages among the workers.
In simple words: We find how much each worker's wage differs from an average wage, then use these differences to calculate the spread (variance) and average spread (standard deviation) of all the wages. This shows how consistent or varied the workers' earnings are.

🎯 Exam Tip: When using the assumed mean method, ensure you calculate \( d_i \) as \( x_i - A \) and then sum both \( d_i \) and \( d_i^2 \) correctly. A common mistake is miscalculating \( (\Sigma d_i/n)^2 \).

Question 6. A wall clock strikes the bell once at 1 o'clock, 2 times at 2 o'clock, 3 times at 3 o'clock and so on. How many times will it strike in a particular day? Find the Standard deviation of the number of strikes the bell make a day.
Answer: First, let's find the total number of strikes in a day.
A wall clock strikes according to the hour in a 12-hour cycle.
Number of strikes in 12 hours \( = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = \frac{12 \times (12+1)}{2} = \frac{12 \times 13}{2} = 6 \times 13 = 78 \) times.
A day has 24 hours, so the clock repeats the 12-hour cycle twice.
Total strikes in a day \( = 2 \times 78 = 156 \) times.

Now, let's find the standard deviation of the number of strikes in a day.
The number of strikes in a day would follow the pattern: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 (this represents the number of strikes if the clock strikes for each hour of a 24-hour cycle, but the source implies an even number of strikes for each hour). Let's follow the solution's interpretation where \( x_i \) values are \( \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \). There are \( n = 12 \) such readings (one for each even hour up to 24).
Assumed mean \( (A) = 14 \).

Answer: (Continued)
Sum of deviations: \( \Sigma d_i = -12 - 10 - 8 - 6 - 4 - 2 + 0 + 2 + 4 + 6 + 8 + 10 = -12 \)
Sum of squared deviations: \( \Sigma d_i^2 = 144 + 100 + 64 + 36 + 16 + 4 + 0 + 4 + 16 + 36 + 64 + 100 = 584 \)
Here \( n = 12 \), \( \Sigma d_i = -12 \), \( \Sigma d_i^2 = 584 \).
The formula for Standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \)
\( \implies \sigma = \sqrt{\frac{584}{12} - \left(\frac{-12}{12}\right)^2} \)
\( \implies \sigma = \sqrt{48.666... - (-1)^2} \)
\( \implies \sigma = \sqrt{48.666... - 1} \)
\( \implies \sigma = \sqrt{47.666...} \)
\( \implies \sigma \approx 6.904 \approx 6.9 \)
The standard deviation of the bell strikes in a day is \( 6.9 \).

**Aliter Method:**
For the data representing strikes at 2, 4, ..., 24 (which can be written as \( 2 \times [1, 2, 3, ..., 12] \)), we can use the formula for standard deviation of an arithmetic progression scaled by a factor.
The standard deviation for the first 'n' natural numbers is \( \text{S.D.} = \sqrt{\frac{n^2-1}{12}} \).
Here, the numbers are \( 2 \times 1, 2 \times 2, ..., 2 \times 12 \). So, \( n=12 \) for the natural numbers inside the bracket, and the data is scaled by 2.
Standard deviation \( (\sigma) = 2 \times \sqrt{\frac{12^2 - 1}{12}} \)
\( \implies \sigma = 2 \times \sqrt{\frac{144 - 1}{12}} \)
\( \implies \sigma = 2 \times \sqrt{\frac{143}{12}} \)
\( \implies \sigma = 2 \times \sqrt{11.9166} \)
\( \implies \sigma = 2 \times 3.45 \)
\( \implies \sigma = 6.9 \)
In simple words: First, sum up the total number of times the clock strikes by counting for 12 hours and doubling it for a full day. Then, to find the standard deviation, we list the number of strikes for each hour in a day, calculate how much each strike count differs from the average, and use a formula to find the overall spread. An easier way for evenly spaced data is to use a direct formula for scaled natural numbers.

🎯 Exam Tip: When a question asks for both a total and a statistical measure, ensure both parts are answered. For patterned data, consider if a shortcut formula for arithmetic progressions can simplify calculations.

Question 7. Find the standard deviation of the first 21 natural numbers.
Answer: We need to find the standard deviation for the set of numbers \( \{1, 2, 3, ..., 21\} \).
Here, \( n = 21 \).
The standard deviation \( (\sigma) \) for the first \( n \) natural numbers is given by the formula:
\( \sigma = \sqrt{\frac{n^2 - 1}{12}} \)
Substitute \( n=21 \):
\( \implies \sigma = \sqrt{\frac{21^2 - 1}{12}} \)
\( \implies \sigma = \sqrt{\frac{441 - 1}{12}} \)
\( \implies \sigma = \sqrt{\frac{440}{12}} \)
\( \implies \sigma = \sqrt{36.666...} \)
\( \implies \sigma \approx 6.055 \approx 6.06 \)
The standard deviation of the first 21 natural numbers is approximately \( 6.06 \). This value indicates the average spread of these numbers around their mean.
In simple words: For a list of numbers starting from 1 and going up to 'n', there's a quick formula to find how much they typically spread out. Just put 'n' into the formula: square 'n', subtract 1, divide by 12, then find the square root.

🎯 Exam Tip: Remember the specific formula for the standard deviation of the first 'n' natural numbers; it saves a lot of calculation time compared to the general formula.

Question 8. If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Answer: The standard deviation is a measure of the spread or dispersion of data points around the mean. It is not affected by adding or subtracting a constant value from each data point.
If each value of the data is decreased by 5, the entire distribution shifts, but its spread remains the same.
Therefore, the new standard deviation will be the same as the original standard deviation.
New standard deviation \( = 4.5 \). This property of standard deviation makes it robust to shifts in data location.
In simple words: When you subtract the same number from every item in a list, the average changes, but how much the numbers are spread out stays exactly the same. So, the standard deviation doesn't change.

🎯 Exam Tip: Know the properties of standard deviation and variance: they are unaffected by shifts in origin (addition/subtraction) but are affected by changes in scale (multiplication/division).

Question 9. If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer: We are given the original standard deviation \( (\sigma) = 3.6 \).
Each value of the data is divided by 3.
When each value in a data set is divided by a constant \( k \), the new standard deviation is the original standard deviation divided by \( k \).
New standard deviation \( (\sigma') = \frac{\text{Original Standard Deviation}}{3} = \frac{3.6}{3} = 1.2 \).
Variance is the square of the standard deviation.
New Variance \( (\sigma'^2) = (\text{New Standard Deviation})^2 = (1.2)^2 = 1.44 \).
Thus, the new standard deviation is \( 1.2 \) and the new variance is \( 1.44 \). This shows how scaling affects the spread of data.
In simple words: If you divide every number in a list by another number, the standard deviation also gets divided by that same number. The variance, which is the standard deviation squared, will then be divided by that number squared.

🎯 Exam Tip: When scaling data, remember that standard deviation scales linearly, while variance scales by the square of the scaling factor.

Question 10. The rainfall recorded in various places of five districts in a week are given below. Find its standard deviation.

Answer: We will use the assumed mean method for calculating standard deviation of grouped discrete data.
Let Rainfall \( x_i \) and Number of places \( f_i \).
Assumed mean \( (A) = 60 \).

Answer: (Continued)
From the table, we have:
\( N = \Sigma f_i = 40 \)
\( \Sigma f_i d_i = -160 \)
\( \Sigma f_i d_i^2 = 3050 \)
The formula for Standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma f_i d_i^2}{N} - \left(\frac{\Sigma f_i d_i}{N}\right)^2} \)
\( \implies \sigma = \sqrt{\frac{3050}{40} - \left(\frac{-160}{40}\right)^2} \)
\( \implies \sigma = \sqrt{76.25 - (-4)^2} \)
\( \implies \sigma = \sqrt{76.25 - 16} \)
\( \implies \sigma = \sqrt{60.25} \)
\( \implies \sigma \approx 7.76 \)
The standard deviation of the rainfall recorded is approximately \( 7.76 \). This indicates how much the rainfall varies across different places.
In simple words: To find how much rainfall varies, we use a table to organize the data by finding the difference from an assumed average. Then, we use these differences in a formula to calculate the standard deviation, which shows the average spread of rainfall amounts.

🎯 Exam Tip: For grouped frequency data, ensure accurate calculation of \( f_i d_i \) and \( f_i d_i^2 \) and their sums, as small errors here can significantly affect the final standard deviation.

Question 11. In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Answer: We will use the assumed mean method for calculating standard deviation for grouped continuous data.
First, find the mid-value \( (x_i) \) for each class interval.
Assumed mean \( (A) = 35 \) (mid-value of the 30-40 class interval).

Answer: (Continued)
From the table, we have:
\( N = \Sigma f_i = 65 \)
\( \Sigma f_i d_i = 30 \)
\( \Sigma f_i d_i^2 = 13900 \)
The formula for Standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma f_i d_i^2}{N} - \left(\frac{\Sigma f_i d_i}{N}\right)^2} \)
\( \implies \sigma = \sqrt{\frac{13900}{65} - \left(\frac{30}{65}\right)^2} \)
\( \implies \sigma = \sqrt{213.846... - (0.4615...)^2} \)
\( \implies \sigma = \sqrt{213.846... - 0.213...} \)
\( \implies \sigma = \sqrt{213.633...} \)
\( \implies \sigma \approx 14.616 \approx 14.62 \)
The standard deviation of the number of people affected is approximately \( 14.62 \). This value helps understand the spread of viral fever across different age groups.
In simple words: To find how much the ages of people affected by fever spread out, we first find the middle age for each group. Then, using an assumed average, we calculate the differences and use these in a formula to get the standard deviation.

🎯 Exam Tip: For continuous grouped data, remember to calculate the mid-value of each class interval accurately, as this forms the \( x_i \) for your calculations.

Question 12. The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Answer: We will use the assumed mean method for calculating standard deviation for grouped continuous data.
First, find the mid-value \( (x_i) \) for each class interval.
Let's adjust the class intervals to be continuous by taking the midpoint between the upper limit of one and the lower limit of the next for accurate mid-values (e.g., for 21-24 and 25-28, the boundary is 24.5). The mid-values would be \( (21+24)/2=22.5 \), \( (25+28)/2=26.5 \), etc.
Assumed mean \( (A) = 34.5 \) (mid-value of the 33-36 class interval).

Answer: (Continued)
From the table, we have:
\( N = \Sigma f_i = 84 \)
\( \Sigma f_i d_i = -316 \)
\( \Sigma f_i d_i^2 = 4208 \)
The formula for Standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma f_i d_i^2}{N} - \left(\frac{\Sigma f_i d_i}{N}\right)^2} \)
\( \implies \sigma = \sqrt{\frac{4208}{84} - \left(\frac{-316}{84}\right)^2} \)
\( \implies \sigma = \sqrt{50.0952... - (-3.7619...)^2} \)
\( \implies \sigma = \sqrt{50.0952... - 14.1520...} \)
\( \implies \sigma = \sqrt{35.943...} \)
\( \implies \sigma \approx 5.995 \approx 6 \)
The standard deviation of the diameters of the plates is approximately \( 6 \) cm. This tells us about the consistency of the plate sizes produced.
In simple words: To find how much the plate diameters vary, we first calculate the middle diameter for each size group. Then, using an estimated average, we find the differences and use these values in a formula to calculate the standard deviation, showing the typical spread of plate sizes.

🎯 Exam Tip: For discontinuous class intervals like 21-24, 25-28, it's good practice to determine the class boundaries (e.g., 20.5-24.5, 24.5-28.5) before finding mid-values, although in this case, direct mid-values \( (21+24)/2 \) are used by the source. Always be consistent with your chosen method.

Question 13. The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Answer: We will use the assumed mean method for calculating standard deviation for grouped continuous data.
First, find the mid-value \( (x_i) \) for each class interval.
Assumed mean \( (A) = 11 \) (mid-value of the 10.5-11.5 class interval).

Answer: (Continued)
From the table, we have:
\( N = \Sigma f_i = 50 \)
\( \Sigma f_i d_i = 8 \)
\( \Sigma f_i d_i^2 = 78 \)
The formula for Standard deviation \( (\sigma) \) is:
\( \sigma = \sqrt{\frac{\Sigma f_i d_i^2}{N} - \left(\frac{\Sigma f_i d_i}{N}\right)^2} \)
\( \implies \sigma = \sqrt{\frac{78}{50} - \left(\frac{8}{50}\right)^2} \)
\( \implies \sigma = \sqrt{1.56 - (0.16)^2} \)
\( \implies \sigma = \sqrt{1.56 - 0.0256} \)
\( \implies \sigma = \sqrt{1.5344} \)
\( \implies \sigma \approx 1.238 \approx 1.24 \)
The standard deviation of the time taken by students is approximately \( 1.24 \) seconds. This measures the consistency of student performance in the race.
In simple words: To find how much the students' race times vary, we calculate the middle time for each group. Then, using an assumed average, we find the differences and use these in a formula to get the standard deviation. This shows how close or far apart the students' finishing times are from the average.

🎯 Exam Tip: For continuous class intervals that are already consecutive (like 8.5-9.5, 9.5-10.5), the mid-values are straightforward. Always double-check your arithmetic, especially for squaring decimals.

Question 14. For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer: We are given the following initial information:
Number of candidates \( (n) = 100 \)
Incorrect Mean \( (\overline{x}) = 60 \)
Incorrect Standard Deviation \( (\sigma) = 15 \)

First, find the incorrect sum of marks \( (\Sigma x) \):
Mean \( = \frac{\Sigma x}{n} \implies 60 = \frac{\Sigma x}{100} \)
\( \implies \Sigma x = 60 \times 100 = 6000 \)

Next, find the incorrect sum of squares \( (\Sigma x^2) \):
Standard Deviation \( (\sigma) = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} \)
\( \implies 15 = \sqrt{\frac{\Sigma x^2}{100} - (60)^2} \)
Squaring both sides:
\( 15^2 = \frac{\Sigma x^2}{100} - 3600 \)
\( 225 = \frac{\Sigma x^2}{100} - 3600 \)
\( 225 + 3600 = \frac{\Sigma x^2}{100} \)
\( 3825 = \frac{\Sigma x^2}{100} \)
\( \implies \Sigma x^2 = 3825 \times 100 = 382500 \)

Now, correct the sum of marks and sum of squares:
Old incorrect values: 40, 27
New correct values: 45, 72

Corrected \( \Sigma x = \text{Incorrect } \Sigma x - (\text{Sum of wrong values}) + (\text{Sum of correct values}) \)
Corrected \( \Sigma x = 6000 - (40 + 27) + (45 + 72) \)
Corrected \( \Sigma x = 6000 - 67 + 117 \)
Corrected \( \Sigma x = 6000 + 50 = 6050 \)

Corrected \( \Sigma x^2 = \text{Incorrect } \Sigma x^2 - (\text{Sum of wrong values squared}) + (\text{Sum of correct values squared}) \)
Corrected \( \Sigma x^2 = 382500 - (40^2 + 27^2) + (45^2 + 72^2) \)
Corrected \( \Sigma x^2 = 382500 - (1600 + 729) + (2025 + 5184) \)
Corrected \( \Sigma x^2 = 382500 - 2329 + 7209 \)
Corrected \( \Sigma x^2 = 380171 + 7209 = 387380 \)

Finally, calculate the corrected mean and standard deviation:
Corrected Mean \( (\overline{x}_{\text{corr}}) = \frac{\text{Corrected } \Sigma x}{n} = \frac{6050}{100} = 60.5 \)

Corrected Standard Deviation \( (\sigma_{\text{corr}}) = \sqrt{\frac{\text{Corrected } \Sigma x^2}{n} - \left(\frac{\text{Corrected } \Sigma x}{n}\right)^2} \)
\( \implies \sigma_{\text{corr}} = \sqrt{\frac{387380}{100} - (60.5)^2} \)
\( \implies \sigma_{\text{corr}} = \sqrt{3873.8 - 3660.25} \)
\( \implies \sigma_{\text{corr}} = \sqrt{213.55} \)
\( \implies \sigma_{\text{corr}} \approx 14.613 \approx 14.61 \)
The corrected mean is \( 60.5 \) and the corrected standard deviation is \( 14.61 \). This process ensures the statistical measures accurately reflect the true data.
In simple words: If you find out some data points were written down wrong, you first figure out the total sum and total sum of squares from the incorrect numbers. Then, you subtract the wrong numbers and add the correct ones to these sums. Finally, you use these new, correct sums to calculate the true average and spread of the data.

🎯 Exam Tip: Correcting mean and standard deviation involves meticulously updating both the sum of observations and the sum of their squares. Always calculate \( \Sigma x \) and \( \Sigma x^2 \) first before applying corrections.

Question 15. The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer: Let the two remaining observations be \( a \) and \( b \).
The seven observations are: \( \{2, 4, 10, 12, 14, a, b\} \).
Number of observations \( (n) = 7 \).
Given Mean \( (\overline{x}) = 8 \).
Given Variance \( (\sigma^2) = 16 \).

First, let's find the sum of all observations \( (\Sigma x_i) \):
\( \overline{x} = \frac{\Sigma x_i}{n} \implies 8 = \frac{2+4+10+12+14+a+b}{7} \)
\( 8 = \frac{42+a+b}{7} \)
\( 8 \times 7 = 42+a+b \)
\( 56 = 42+a+b \)
\( a+b = 56 - 42 \)
\( \implies a+b = 14 \) ... (1)

Next, let's find the sum of squares of all observations \( (\Sigma x_i^2) \):
The values for \( x_i \) and \( x_i^2 \) are as follows:

Answer: (Continued)
\( \Sigma x_i = 2+4+10+12+14+a+b = 42+a+b \)
\( \Sigma x_i^2 = 4+16+100+144+196+a^2+b^2 = 460+a^2+b^2 \)

The formula for Variance \( (\sigma^2) \) is:
\( \sigma^2 = \frac{\Sigma x_i^2}{n} - \left(\frac{\Sigma x_i}{n}\right)^2 \)
We know \( \frac{\Sigma x_i}{n} = \overline{x} = 8 \).
\( 16 = \frac{460+a^2+b^2}{7} - (8)^2 \)
\( 16 = \frac{460+a^2+b^2}{7} - 64 \)
\( 16 + 64 = \frac{460+a^2+b^2}{7} \)
\( 80 = \frac{460+a^2+b^2}{7} \)
\( 80 \times 7 = 460+a^2+b^2 \)
\( 560 = 460+a^2+b^2 \)
\( a^2+b^2 = 560 - 460 \)
\( \implies a^2+b^2 = 100 \) ... (2)

Now we have a system of two equations:
1. \( a+b = 14 \)
2. \( a^2+b^2 = 100 \)

From equation (1), \( a = 14-b \). Substitute this into equation (2):
\( (14-b)^2 + b^2 = 100 \)
\( 196 - 28b + b^2 + b^2 = 100 \)
\( 2b^2 - 28b + 196 - 100 = 0 \)
\( 2b^2 - 28b + 96 = 0 \)
Divide by 2:
\( b^2 - 14b + 48 = 0 \)
Factor the quadratic equation:
\( (b-6)(b-8) = 0 \)
So, \( b=6 \) or \( b=8 \).

If \( b=6 \), then from \( a+b=14 \), \( a = 14-6 = 8 \).
If \( b=8 \), then from \( a+b=14 \), \( a = 14-8 = 6 \).
Thus, the two remaining observations are 6 and 8. These missing values complete the dataset and allow for consistent statistical analysis.
In simple words: When some numbers are missing from a list, but you know the average (mean) and how spread out they are (variance), you can use these facts to find the missing numbers. This often involves setting up and solving simple algebra problems.

🎯 Exam Tip: This problem often leads to a system of quadratic equations. Remember the identity \( (a+b)^2 = a^2+b^2+2ab \) or \( a^2+b^2 = (a+b)^2 - 2ab \) to simplify the algebraic solution.