Samacheer Kalvi Class 10 Maths Solutions Chapter 8 Statistics and Probability Exercise 8.2

 

Question 1. The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer: The standard deviation (\( \sigma \)) of the data is 6.5.
The mean (\( \bar{x} \)) of the data is 12.5.
We use the formula for the coefficient of variation:
\( \text{Coefficient of variation} = \frac{\sigma}{\bar{x}} \times 100 \% \)
Substitute the given values:
\( \text{Coefficient of variation} = \frac{6.5}{12.5} \times 100 \% \)
\( \implies \text{Coefficient of variation} = 0.52 \times 100 \% \)
\( \implies \text{Coefficient of variation} = 52 \% \)
The coefficient of variation helps us understand how much the data spreads out compared to its average.
In simple words: We divide the standard deviation (how spread out the numbers are) by the mean (the average) and then multiply by 100 to get the percentage. This shows us the relative spread of the data.

🎯 Exam Tip: Remember to correctly identify the standard deviation (\( \sigma \)) and the mean (\( \bar{x} \)) from the problem statement before applying the formula.

 

Question 2. The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer: The standard deviation (\( \sigma \)) of the data is 1.2.
The coefficient of variation (C.V.) is 25.6.
We use the formula for the coefficient of variation:
\( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)
Substitute the known values into the formula:
\( 25.6 = \frac{1.2}{\bar{x}} \times 100 \)
To find the mean (\( \bar{x} \)), we rearrange the equation:
\( 25.6 \bar{x} = 1.2 \times 100 \)
\( \implies 25.6 \bar{x} = 120 \)
\( \implies \bar{x} = \frac{120}{25.6} \)
\( \implies \bar{x} = 4.6875 \)
Rounding to two decimal places, the value of the mean is approximately 4.69. The mean is a measure of the central tendency of the data.
In simple words: We use the formula for coefficient of variation, put in the numbers we know, and then do some algebra to find the missing mean. The mean tells us the average value.

🎯 Exam Tip: When solving for a missing variable like the mean, make sure to isolate it correctly using algebraic rearrangement steps. Be careful with decimal division.

 

Question 3. If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Answer: The mean (\( \bar{x} \)) of the data is 15.
The coefficient of variation (C.V.) is 48.
We use the formula for the coefficient of variation:
\( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)
Substitute the given values into the formula:
\( 48 = \frac{\sigma}{15} \times 100 \)
To find the standard deviation (\( \sigma \)), we rearrange the equation:
\( 48 \times 15 = \sigma \times 100 \)
\( \implies 720 = 100 \sigma \)
\( \implies \sigma = \frac{720}{100} \)
\( \implies \sigma = 7.2 \)
The standard deviation shows how much individual data points typically vary from the mean.
In simple words: We have the average and the coefficient of variation. By using the formula, we can work backwards to find out how spread out the data points are, which is the standard deviation.

🎯 Exam Tip: Always double-check your multiplication and division when rearranging the formula to avoid calculation errors.

 

Question 4. If n = 5, \( \bar{x} = 6 \), \( \Sigma x^2 = 765 \), then calculate the coefficient of variation.
Answer: Given:
Number of observations \( n = 5 \)
Mean \( \bar{x} = 6 \)
Sum of squares \( \Sigma x^2 = 765 \)
We first need to calculate the standard deviation (\( \sigma \)) using the formula:
\( \sigma = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x})^2} \)
Substitute the given values:
\( \sigma = \sqrt{\frac{765}{5} - (6)^2} \)
\( \implies \sigma = \sqrt{153 - 36} \)
\( \implies \sigma = \sqrt{117} \)
\( \implies \sigma \approx 10.8166 \)
Now, we calculate the coefficient of variation (C.V.) using the formula:
\( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)
Substitute the values of \( \sigma \) and \( \bar{x} \):
\( \text{C.V.} = \frac{10.8166}{6} \times 100 \% \)
\( \implies \text{C.V.} \approx 1.80276 \times 100 \% \)
\( \implies \text{C.V.} \approx 180.27 \% \)
This high coefficient of variation shows that the data points are very spread out compared to their mean.
In simple words: First, we find how spread out the numbers are using the standard deviation formula. Then, we use that answer along with the average to calculate the coefficient of variation, which tells us how much the numbers vary relative to their average.

🎯 Exam Tip: Ensure you use the correct formula for standard deviation based on the given information (raw data sum of squares and mean). Rounding intermediate steps too early can affect the final answer.

 

Question 5. Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Answer: Given data values are 24, 26, 33, 37, 29, 31.
First, arrange the data in ascending order: 24, 26, 29, 31, 33, 37.
The number of data points \( n = 6 \).
We will use the assumed mean method for calculations. Let the assumed mean \( A = 29 \).
Now, we create a table to calculate \( d_i = x_i - A \) and \( d_i^2 \):

Now, we calculate the mean (\( \bar{x} \)):
\( \bar{x} = A + \frac{\Sigma d_i}{n} = 29 + \frac{6}{6} = 29 + 1 = 30 \)
Next, we calculate the standard deviation (\( \sigma \)):
\( \sigma = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \)
Substitute the values from the table:
\( \sigma = \sqrt{\frac{118}{6} - \left(\frac{6}{6}\right)^2} \)
\( \implies \sigma = \sqrt{19.666... - (1)^2} \)
\( \implies \sigma = \sqrt{19.666... - 1} \)
\( \implies \sigma = \sqrt{18.666...} \)
\( \implies \sigma \approx 4.3204 \)
Finally, we calculate the coefficient of variation (C.V.):
\( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)
\( \text{C.V.} = \frac{4.3204}{30} \times 100 \% \)
\( \implies \text{C.V.} \approx 0.14401 \times 100 \% \)
\( \implies \text{C.V.} \approx 14.40 \% \)
The coefficient of variation gives a clear picture of how much the data values vary relative to their average value.
In simple words: We first find the average of the numbers. Then, we find how spread out they are using the standard deviation. Finally, we use both these values to calculate the coefficient of variation, which shows us the percentage of variability in the data.

🎯 Exam Tip: When given a list of raw data, carefully calculate the mean and standard deviation. Using an assumed mean can simplify calculations, but ensure you apply the correct formula adjustments.

 

Question 6. The time taken (in minutes) to complete homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Answer: Given times (in minutes) for 8 students: 38, 40, 47, 44, 46, 43, 49, 53.
First, arrange the data in ascending order: 38, 40, 43, 44, 46, 47, 49, 53.
The number of data points \( n = 8 \).
We use the assumed mean method for calculations. Let the assumed mean \( A = 46 \).
Now, we create a table to calculate \( d_i = x_i - A \) and \( d_i^2 \):

Now, we calculate the mean (\( \bar{x} \)):
\( \bar{x} = A + \frac{\Sigma d_i}{n} = 46 + \frac{-8}{8} = 46 - 1 = 45 \)
Next, we calculate the standard deviation (\( \sigma \)):
\( \sigma = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \)
Substitute the values from the table:
\( \sigma = \sqrt{\frac{172}{8} - \left(\frac{-8}{8}\right)^2} \)
\( \implies \sigma = \sqrt{21.5 - (-1)^2} \)
\( \implies \sigma = \sqrt{21.5 - 1} \)
\( \implies \sigma = \sqrt{20.5} \)
\( \implies \sigma \approx 4.5276 \)
Finally, we calculate the coefficient of variation (C.V.):
\( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)
\( \text{C.V.} = \frac{4.5276}{45} \times 100 \% \)
\( \implies \text{C.V.} \approx 0.10061 \times 100 \% \)
\( \implies \text{C.V.} \approx 10.06 \% \)
A lower coefficient of variation here indicates more consistent homework completion times among students.
In simple words: We find the average time students spent on homework. Then we calculate how much those times vary. We put these two numbers into a formula to get the coefficient of variation, which tells us how spread out the homework times are as a percentage.

🎯 Exam Tip: Be careful with negative signs when calculating \( d_i \) and \( (\frac{\Sigma d_i}{n})^2 \). Squaring a negative number always results in a positive number.

 

Question 7. The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation of 4.6 and 2.4 respectively. Who is more consistent in performance?
Answer: To determine who is more consistent, we need to calculate the coefficient of variation (C.V.) for both Sathya and Vidhya. The student with a lower C.V. is more consistent.
Given:
Number of subjects \( n = 5 \)

**For Sathya:**
Total marks scored = 460
Standard deviation \( \sigma_{\text{Sathya}} = 4.6 \)
Calculate the mean marks for Sathya:
\( \bar{x}_{\text{Sathya}} = \frac{\text{Total marks}}{n} = \frac{460}{5} = 92 \)
Calculate the coefficient of variation for Sathya:
\( \text{C.V.}_{\text{Sathya}} = \frac{\sigma_{\text{Sathya}}}{\bar{x}_{\text{Sathya}}} \times 100 \% \)
\( \text{C.V.}_{\text{Sathya}} = \frac{4.6}{92} \times 100 \% \)
\( \implies \text{C.V.}_{\text{Sathya}} = 0.05 \times 100 \% \)
\( \implies \text{C.V.}_{\text{Sathya}} = 5 \% \)

**For Vidhya:**
Total marks scored = 480
Standard deviation \( \sigma_{\text{Vidhya}} = 2.4 \)
Calculate the mean marks for Vidhya:
\( \bar{x}_{\text{Vidhya}} = \frac{\text{Total marks}}{n} = \frac{480}{5} = 96 \)
Calculate the coefficient of variation for Vidhya:
\( \text{C.V.}_{\text{Vidhya}} = \frac{\sigma_{\text{Vidhya}}}{\bar{x}_{\text{Vidhya}}} \times 100 \% \)
\( \text{C.V.}_{\text{Vidhya}} = \frac{2.4}{96} \times 100 \% \)
\( \implies \text{C.V.}_{\text{Vidhya}} = 0.025 \times 100 \% \)
\( \implies \text{C.V.}_{\text{Vidhya}} = 2.5 \% \)

**Comparison:**
Since \( \text{C.V.}_{\text{Vidhya}} (2.5 \%) < \text{C.V.}_{\text{Sathya}} (5 \%) \), Vidhya is more consistent in her performance. A lower coefficient of variation indicates greater consistency.
In simple words: We found the average marks and how much marks changed for each student. Vidhya's marks changed less compared to her average, which means she was more consistent in her school performance.

🎯 Exam Tip: Consistency is directly related to the coefficient of variation. A smaller C.V. always means greater consistency in the data set.

 

Question 8. The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below. Which of the three subjects shows the highest variation and which shows the lowest variation in marks?
Answer: To find the subject with the highest and lowest variation, we need to calculate the coefficient of variation (C.V.) for each subject. A higher C.V. means more variation, and a lower C.V. means less variation.
Given data:

We use the formula: \( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)

(i) **For Mathematics:**
Mean \( \bar{x} = 56 \)
Standard deviation \( \sigma = 12 \)
\( \text{C.V.}_1 = \frac{12}{56} \times 100 \% \)
\( \implies \text{C.V.}_1 \approx 0.214285 \times 100 \% \)
\( \implies \text{C.V.}_1 \approx 21.43 \% \)

(ii) **For Science:**
Mean \( \bar{x} = 65 \)
Standard deviation \( \sigma = 14 \)
\( \text{C.V.}_2 = \frac{14}{65} \times 100 \% \)
\( \implies \text{C.V.}_2 \approx 0.215384 \times 100 \% \)
\( \implies \text{C.V.}_2 \approx 21.54 \% \)

(iii) **For Social Science:**
Mean \( \bar{x} = 60 \)
Standard deviation \( \sigma = 10 \)
\( \text{C.V.}_3 = \frac{10}{60} \times 100 \% \)
\( \implies \text{C.V.}_3 \approx 0.166666 \times 100 \% \)
\( \implies \text{C.V.}_3 \approx 16.67 \% \)

**Comparison of C.V. values:**
Mathematics: 21.43 %
Science: 21.54 %
Social Science: 16.67 %

From the calculations:
* Science has the highest coefficient of variation (21.54 %), meaning it shows the **highest variation** in marks.
* Social Science has the lowest coefficient of variation (16.67 %), meaning it shows the **lowest variation** in marks.
This tells us that students' scores are most spread out in Science and most consistent in Social Science.
In simple words: We calculate a special percentage for each subject that tells us how much the student scores change. The subject with the biggest percentage has the most different scores, and the one with the smallest percentage has scores that are very similar.

🎯 Exam Tip: To compare variation between different datasets, always calculate the coefficient of variation, as it is a relative measure and is not affected by the scale of the data.

 

Question 9. The temperature of two cities A and B in the winter season are given below. Find which city is more consistent in temperature changes?
Answer: To find which city is more consistent in temperature changes, we need to calculate the coefficient of variation (C.V.) for both City A and City B. The city with a lower C.V. will have more consistent temperatures.
Given temperatures:

We use the formula: \( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \% \)

(i) **For City A:**
Data: 18, 20, 22, 24, 26. Number of observations \( n = 5 \).
Let assumed mean \( A = 22 \).Mean \( \bar{x}_A = A + \frac{\Sigma d_i}{n} = 22 + \frac{0}{5} = 22 \)
Standard deviation \( \sigma_A = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \)
\( \sigma_A = \sqrt{\frac{40}{5} - \left(\frac{0}{5}\right)^2} \)
\( \implies \sigma_A = \sqrt{8 - 0} = \sqrt{8} \)
\( \implies \sigma_A \approx 2.8284 \)
Coefficient of variation \( \text{C.V.}_A = \frac{2.8284}{22} \times 100 \% \)
\( \implies \text{C.V.}_A \approx 12.856 \% \approx 12.86 \% \)

(ii) **For City B:**
Data: 11, 14, 15, 17, 18. Number of observations \( n = 5 \).
Let assumed mean \( A = 15 \).Mean \( \bar{x}_B = A + \frac{\Sigma d_i}{n} = 15 + \frac{0}{5} = 15 \)
Standard deviation \( \sigma_B = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \)
\( \sigma_B = \sqrt{\frac{30}{5} - \left(\frac{0}{5}\right)^2} \)
\( \implies \sigma_B = \sqrt{6 - 0} = \sqrt{6} \)
\( \implies \sigma_B \approx 2.4495 \)
Coefficient of variation \( \text{C.V.}_B = \frac{2.4495}{15} \times 100 \% \)
\( \implies \text{C.V.}_B \approx 16.33 \% \)

**Comparison:**
Since \( \text{C.V.}_A (12.86 \%) < \text{C.V.}_B (16.33 \%) \), City A is more consistent in its temperature changes. This means temperatures in City A are more stable and fluctuate less relative to their average compared to City B.
In simple words: We calculated how much the temperature changes in each city compared to its average temperature. City A had a smaller percentage change, which means its temperature was more steady and predictable than City B's.

🎯 Exam Tip: Always remember that a lower coefficient of variation signifies more consistency, while a higher value indicates greater variability. Clearly state your conclusion based on this principle.