Dispersion: 17 flashcards
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1. If all observations in a data set are equal ($x_1 = x_2 = \ldots = x_n = k$), what is the value of any measure of dispersion?
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Zero. There is no spread when every observation equals the same constant $k$.
2. Define standard deviation for a set of $n$ observations $x_1, x_2, \ldots, x_n$ with mean $\bar{x}$.
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$$s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n}}$$
3. What is the formula for the coefficient of standard deviation?
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$$\text{Coefficient of SD} = \frac{s}{\bar{x}}$$ where $s$ is the standard deviation and $\bar{x}$ is the mean.
4. What is the formula for the coefficient of variation (CV)?
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$$\text{CV} = \frac{s}{\bar{x}} \times 100\%$$
5. For grouped data with class marks $x_i$ and frequencies $f_i$, write the formula for standard deviation.
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$$s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}}$$ where $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
6. For a frequency distribution, how do you find the class mark of a class interval?
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Class mark $= \frac{\text{Lower limit} + \text{Upper limit}}{2}$.
7. Two batsmen A and B have CV of runs 20% and 35% respectively. Who is more consistent?
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Batsman A is more consistent. A smaller coefficient of variation means less relative dispersion and greater consistency.
8. Company A has mean salary ₹600 and SD ₹30; Company B has mean salary ₹2100 and SD ₹84. Which company has more stable salaries?
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Company A. CV$_A = \frac{30}{600} \times 100 = 5\%$; CV$_B = \frac{84}{2100} \times 100 = 4\%$. Company B actually has lower CV, so **Company B** is more stable. (Recompute carefully: $84/2100 = 0.04 = 4\%$.)
9. The CVs of two series are 30% and 25%, with standard deviations 15 and 9 respectively. Find their means.
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Series 1: $\bar{x}_1 = \frac{15}{0.30} = 50$. Series 2: $\bar{x}_2 = \frac{9}{0.25} = 36$.
10. For two groups $G_1$ and $G_2$ combined into group $G$, write the formula for the combined mean $\bar{x}$.
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$$\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$
11. State the formula for the combined standard deviation of two groups $G_1$ and $G_2$.
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$$s = \sqrt{\frac{n_1(s_1^2 + d_1^2) + n_2(s_2^2 + d_2^2)}{n_1 + n_2}}$$ where $d_1 = \bar{x}_1 - \bar{x}$ and $d_2 = \bar{x}_2 - \bar{x}$.
12. Marks of 9 students: 64, 63, 72, 65, 68, 69, 66, 67, 69. Find the standard deviation.
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$\bar{x} = 67$, $\sum (x_i - \bar{x})^2 = 54$, so $s = \sqrt{\frac{54}{9}} = \sqrt{6} \approx 2.45$.
13. Number of cars at 5 service stations: 7, 3, 11, 8, 9. Find the standard deviation.
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$\bar{x} = 7.6$, $\sum (x_i - \bar{x})^2 = 34.8$, so $s = \sqrt{\frac{34.8}{5}} = \sqrt{6.96} \approx 2.64$.
14. In a grouped frequency distribution, what does $f_i$ represent?
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$f_i$ is the frequency of the $i$-th class, i.e., the number of observations falling in that class interval.
15. Why is the coefficient of variation preferred over standard deviation when comparing two data sets with different units or very different means?
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CV is a unit-free relative measure ($s/\bar{x} \times 100\%$), so it allows fair comparison of variability across data sets with different scales or units.
16. Deposits (thousand ₹): 5, 10, 15, 20, 25, 30, 35 with frequencies 2, 7, 11, 15, 10, 4, 1. What is $\sum f_i$?
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$\sum f_i = 2 + 7 + 11 + 15 + 10 + 4 + 1 = 50$.
17. For the deposits data (class marks 5 to 35, frequencies 2, 7, 11, 15, 10, 4, 1), compute the mean deposit.
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$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{10 + 70 + 165 + 300 + 250 + 120 + 35}{50} = \frac{950}{50} = 19$ thousand ₹.
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