Analysis of Statistics: Miscellaneous Exercise - Class 11 Maths NCERT

Recap: Key Measures of Dispersion

Before diving into the miscellaneous problems, let's quickly recap the essential measures of dispersion that will be tested. These measures tell us how spread out our data points are from a central value. Understanding their nuances is key to solving diverse problems.

1. Mean Deviation (M.D.): This measures the average of the absolute differences between each data point and either the mean or the median. It's relatively easy to calculate and interpret, giving a simple idea of spread. However, because it ignores the sign of the deviations, it's not as mathematically robust as variance or standard deviation.

• About Mean: Average of $|x_i - \bar{x}|$
• About Median: Average of $|x_i - M|$

1. Variance ($\\sigma^2$): This is the average of the squared differences from the mean. Squaring the deviations makes all values positive and gives more weight to larger deviations, making it more sensitive to outliers. Variance is fundamental in statistical inference and probability theory, but its units are squared, which can be less intuitive.

• For individual data: $\frac{\sum (x_i - \bar{x})^2}{n}$
• For grouped data: $\frac{\sum f_i (x_i - \bar{x})^2}{N}$

1. Standard Deviation ($\\sigma$): The square root of the variance. This brings the measure of dispersion back into the same units as the original data, making it much easier to interpret. It's the most widely used measure of spread and is critical for understanding the distribution of data.

1. Coefficient of Variation (C.V.): This is a relative measure of dispersion, expressed as a percentage. It's calculated as $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100$. C.V. is particularly useful for comparing the consistency or variability of two different data sets, even if they have different means or are measured in different units. A lower C.V. indicates greater consistency.

Important Statistical Terms

Measures of Dispersion

Statistical values that describe the spread or variability of a data set. They indicate how much individual data points differ from the average or central value.

Mean Deviation

The average of the absolute deviations of observations from a measure of central tendency (either the mean or the median).

Variance

The average of the squared differences from the mean. It quantifies the degree of spread in a data set.

Standard Deviation

The square root of the variance. It is the most common measure of dispersion, expressed in the same units as the data, making it easy to interpret.

Coefficient of Variation (C.V.)

A relative measure of dispersion, calculated as the ratio of standard deviation to the mean, usually expressed as a percentage. It's used for comparing variability across different data sets.

Solved Examples for Comprehensive Understanding

• Example 1: Calculating Mean, Variance, and Standard Deviation for Grouped Data Problem: The scores of a batsman in 10 matches are: 38, 70, 48, 34, 42, 55, 63, 46, 54, 44. Find the mean, variance, and standard deviation. Solution: Step 1: Calculate the Mean ($\bar{x}$) Sum of scores ($\sum x_i$) = 38+70+48+34+42+55+63+46+54+44 = 494 Number of matches (n) = 10 Mean ($\bar{x}$) = $\frac{\sum x_i}{n} = \frac{494}{10} = 49.4$ Step 2: Calculate Deviations from the Mean (x_i - \bar{x}) and their Squares (x_i - \bar{x})^2 | $x_i$ | $x_i - \bar{x}$ | $(x_i - \bar{x})^2$ | |---|---|---| | 38 | 38 - 49.4 = -11.4 | 129.96 | | 70 | 70 - 49.4 = 20.6 | 424.36 | | 48 | 48 - 49.4 = -1.4 | 1.96 | | 34 | 34 - 49.4 = -15.4 | 237.16 | | 42 | 42 - 49.4 = -7.4 | 54.76 | | 55 | 55 - 49.4 = 5.6 | 31.36 | | 63 | 63 - 49.4 = 13.6 | 184.96 | | 46 | 46 - 49.4 = -3.4 | 11.56 | | 54 | 54 - 49.4 = 4.6 | 21.16 | | 44 | 44 - 49.4 = -5.4 | 29.16 | Sum of squared deviations ($\sum (x_i - \bar{x})^2$) = 129.96 + 424.36 + 1.96 + 237.16 + 54.76 + 31.36 + 184.96 + 11.56 + 21.16 + 29.16 = 1126.32 Step 3: Calculate Variance ($\sigma^2$) $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{1126.32}{10} = 112.632$ Step 4: Calculate Standard Deviation ($\sigma$) $\sigma = \sqrt{\sigma^2} = \sqrt{112.632} \approx 10.613$ Final Answer: Mean = 49.4, Variance = 112.632, Standard Deviation $\approx$ 10.613. ---
• Example 2: Comparing Consistency using Coefficient of Variation Problem: The scores of two batsmen, A and B, in a series of 10 matches are given below: Batsman A: 10, 15, 80, 70, 12, 11, 14, 16, 18, 100 Batsman B: 30, 40, 35, 45, 50, 60, 25, 55, 38, 42 Which batsman is more consistent? Solution: To compare consistency, we need to calculate the Coefficient of Variation (C.V.) for both batsmen. The batsman with a lower C.V. is more consistent. For Batsman A: Step 1: Calculate Mean ($\bar{x}_A$) $\sum x_A = 10+15+80+70+12+11+14+16+18+100 = 346$ $\bar{x}_A = \frac{346}{10} = 34.6$ Step 2: Calculate Variance ($\sigma_A^2$) We'll use the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ $\sum x_A^2 = 10^2+15^2+80^2+70^2+12^2+11^2+14^2+16^2+18^2+100^2$ $= 100+225+6400+4900+144+121+196+256+324+10000 = 22566$ $\sigma_A^2 = \frac{22566}{10} - (34.6)^2 = 2256.6 - 1197.16 = 1059.44$ Step 3: Calculate Standard Deviation ($\sigma_A$) $\sigma_A = \sqrt{1059.44} \approx 32.55$ Step 4: Calculate Coefficient of Variation (C.V. for A) C.V.$_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{32.55}{34.6} \times 100 \approx 94.07%$ For Batsman B: Step 1: Calculate Mean ($\bar{x}_B$) $\sum x_B = 30+40+35+45+50+60+25+55+38+42 = 420$ $\bar{x}_B = \frac{420}{10} = 42$ Step 2: Calculate Variance ($\sigma_B^2$) $\sum x_B^2 = 30^2+40^2+35^2+45^2+50^2+60^2+25^2+55^2+38^2+42^2$ $= 900+1600+1225+2025+2500+3600+625+3025+1444+1764 = 18708$ $\sigma_B^2 = \frac{18708}{10} - (42)^2 = 1870.8 - 1764 = 106.8$ Step 3: Calculate Standard Deviation ($\sigma_B$) $\sigma_B = \sqrt{106.8} \approx 10.33$ Step 4: Calculate Coefficient of Variation (C.V. for B) C.V.$_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{10.33}{42} \times 100 \approx 24.59%$ Step 5: Compare C.V. Since C.V.$_B$ (24.59%) < C.V.$_A$ (94.07%), Batsman B is more consistent. Final Answer: Batsman B is more consistent.

Exam Tips & Common Mistakes to Avoid

Miscellaneous exercises often combine different concepts, so precision and attention to detail are key. Here are some critical tips to help you ace these problems:

1. Read the Question Carefully: Always identify what type of data you're dealing with (individual, discrete, or continuous/grouped) and which measure of dispersion is being asked for. A common mistake is to use the wrong formula.
2. Accuracy in Calculations: Statistical calculations can be lengthy. Use a calculator diligently and double-check your sums, especially for $\sum x_i$, $\sum f_i$, $\sum x_i^2$, and $\sum f_i x_i^2$. Small arithmetic errors can cascade and lead to incorrect final answers.
3. Correct Mid-points for Grouped Data: When dealing with continuous frequency distributions, ensure you calculate the mid-points ($x_i$) correctly for each class interval. An error here will affect all subsequent calculations.
4. Difference between Mean Deviation and Standard Deviation: Understand that Mean Deviation uses absolute differences ($|x_i - \bar{x}|$), while Variance/Standard Deviation uses squared differences ($(x_i - \bar{x})^2$). This distinction is vital.
5. Understanding Coefficient of Variation: Remember that a lower C.V. indicates greater consistency or lesser variability. This is a common interpretation error. Always relate C.V. back to the context of the problem.
6. Formula for Combined Variance (if applicable): While less common in Class 11, be aware of the formula for combined variance if you've covered it, as miscellaneous problems can test such advanced concepts. However, for most NCERT miscellaneous exercises, focus on the individual measures.
7. Practice: The best way to avoid mistakes is consistent practice. Solve a variety of problems from the textbook and other sources.

Practice Questions with Solutions

• Q: The mean and variance of 7 observations are 8 and 16, respectively. If each observation is multiplied by 3, find the new mean and variance. A: Step 1: Given mean ($\bar{x}$) = 8, variance ($\sigma^2$) = 16, number of observations (n) = 7. Step 2: If each observation is multiplied by a constant 'k' (here k=3), the new mean will be $k \bar{x}$. New mean = $3 \times 8 = 24$. Step 3: If each observation is multiplied by a constant 'k', the new variance will be $k^2 \sigma^2$. New variance = $3^2 \times 16 = 9 \times 16 = 144$. Final answer: The new mean is 24 and the new variance is 144.
• Q: Find the mean deviation about the median for the following data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19. A: Step 1: Arrange the data in ascending order: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19. Step 2: Calculate the median (M). Since there are 10 observations (even), the median is the average of the 5th and 6th observations. M = $\frac{7+9}{2} = \frac{16}{2} = 8$. Step 3: Calculate the absolute deviations from the median ($|x_i - M|$). $|3-8|=5, |3-8|=5, |4-8|=4, |5-8|=3, |7-8|=1, |9-8|=1, |10-8|=2, |12-8|=4, |18-8|=10, |19-8|=11$. Step 4: Sum the absolute deviations: $\sum |x_i - M| = 5+5+4+3+1+1+2+4+10+11 = 46$. Step 5: Calculate the Mean Deviation about Median (M.D. (M)). M.D. (M) = $\frac{\sum |x_i - M|}{n} = \frac{46}{10} = 4.6$. Final answer: The mean deviation about the median is 4.6.
• Q: The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2, and 6, find the other two observations. A: Step 1: Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $n=5$, mean $\bar{x}=4.4$, variance $\sigma^2=8.24$. Let the unknown observations be 'a' and 'b'. The observations are 1, 2, 6, a, b. Step 2: Use the mean formula: $\bar{x} = \frac{\sum x_i}{n}$. $4.4 = \frac{1+2+6+a+b}{5} \Rightarrow 22 = 9+a+b \Rightarrow a+b = 13$ (Equation 1) Step 3: Use the variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $8.24 = \frac{1^2+2^2+6^2+a^2+b^2}{5} - (4.4)^2$ $8.24 = \frac{1+4+36+a^2+b^2}{5} - 19.36$ $8.24 + 19.36 = \frac{41+a^2+b^2}{5}$ $27.6 = \frac{41+a^2+b^2}{5}$ $138 = 41+a^2+b^2 \Rightarrow a^2+b^2 = 97$ (Equation 2) Step 4: From Equation 1, $b = 13 - a$. Substitute this into Equation 2. $a^2 + (13-a)^2 = 97$ $a^2 + 169 - 26a + a^2 = 97$ $2a^2 - 26a + 169 - 97 = 0$ $2a^2 - 26a + 72 = 0$ $a^2 - 13a + 36 = 0$ Step 5: Solve the quadratic equation for 'a'. $(a-4)(a-9) = 0$ So, $a=4$ or $a=9$. Step 6: If $a=4$, then $b=13-4=9$. If $a=9$, then $b=13-9=4$. Final answer: The other two observations are 4 and 9.
• Q: Two factories A and B produced the following items: Factory A: Mean production = 200 units, Standard Deviation = 10 units. Factory B: Mean production = 250 units, Standard Deviation = 12 units. Which factory shows greater consistency in production? A: Step 1: To compare consistency, calculate the Coefficient of Variation (C.V.) for each factory. Lower C.V. means greater consistency. Step 2: For Factory A: Mean ($\bar{x}_A$) = 200 units Standard Deviation ($\sigma_A$) = 10 units C.V.$_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{10}{200} \times 100 = 5%$. Step 3: For Factory B: Mean ($\bar{x}_B$) = 250 units Standard Deviation ($\sigma_B$) = 12 units C.V.$_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{12}{250} \times 100 = 4.8%$. Step 4: Compare C.V. values. Since C.V.$_B$ (4.8%) < C.V.$_A$ (5%), Factory B shows greater consistency. Final answer: Factory B shows greater consistency in production.

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