From the prices of shares $X$ and $Y$ below,find out which is more stable in value:
$X$ $35$ $54$ $52$ $53$ $56$ $58$ $52$ $50$ $51$ $49$ $Y$ $108$ $107$ $105$ $105$ $106$ $107$ $104$ $103$ $104$ $101$
| $X$ | $35$ | $54$ | $52$ | $53$ | $56$ | $58$ | $52$ | $50$ | $51$ | $49$ |
| $Y$ | $108$ | $107$ | $105$ | $105$ | $106$ | $107$ | $104$ | $103$ | $104$ | $101$ |
(B) To determine stability,we calculate the Coefficient of Variation $(C.V.)$. The share with the lower $C.V.$ is more stable.
For share $X$: Data = $35, 54, 52, 53, 56, 58, 52, 50, 51, 49$. Mean $\bar{x} = 51$. Variance $\sigma_1^2 = 35$. Standard deviation $\sigma_1 = \sqrt{35} \approx 5.916$.
$C.V._X = (\sigma_1 / \bar{x}) \times 100 = (5.916 / 51) \times 100 \approx 11.60$.
For share $Y$: Data = $108, 107, 105, 105, 106, 107, 104, 103, 104, 101$. Mean $\bar{y} = 105$. Variance $\sigma_2^2 = 4$. Standard deviation $\sigma_2 = \sqrt{4} = 2$.
$C.V._Y = (\sigma_2 / \bar{y}) \times 100 = (2 / 105) \times 100 \approx 1.90$.
Since $C.V._Y < C.V._X$,share $Y$ is more stable.
For share $X$: Data = $35, 54, 52, 53, 56, 58, 52, 50, 51, 49$. Mean $\bar{x} = 51$. Variance $\sigma_1^2 = 35$. Standard deviation $\sigma_1 = \sqrt{35} \approx 5.916$.
$C.V._X = (\sigma_1 / \bar{x}) \times 100 = (5.916 / 51) \times 100 \approx 11.60$.
For share $Y$: Data = $108, 107, 105, 105, 106, 107, 104, 103, 104, 101$. Mean $\bar{y} = 105$. Variance $\sigma_2^2 = 4$. Standard deviation $\sigma_2 = \sqrt{4} = 2$.
$C.V._Y = (\sigma_2 / \bar{y}) \times 100 = (2 / 105) \times 100 \approx 1.90$.
Since $C.V._Y < C.V._X$,share $Y$ is more stable.
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Similar Questions
Assertion $(A)$: The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.
Reason $(R)$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2-1)}{3}$.
Which of the following alternatives is correct?
Reason $(R)$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2-1)}{3}$.
Which of the following alternatives is correct?
DifficultTS EAMCET 2023
View SolutionThe variance of the following continuous frequency distribution is
$\begin{array}{|l|c|c|c|c|}\hline \text{Class interval} & 0-4 & 4-8 & 8-12 & 12-16 \\ \hline \text{Frequency} & 2 & 3 & 2 & 1 \\ \hline\end{array}$
$\begin{array}{|l|c|c|c|c|}\hline \text{Class interval} & 0-4 & 4-8 & 8-12 & 12-16 \\ \hline \text{Frequency} & 2 & 3 & 2 & 1 \\ \hline\end{array}$
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$ $0$ $1$ $5$ $6$ $10$ $12$ $17$ $f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$
| $x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
| $f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
The variance of the following distribution is:
Marks $1-3$ $3-5$ $5-7$ $7-9$ Number of students $40$ $30$ $20$ $10$
| Marks | $1-3$ | $3-5$ | $5-7$ | $7-9$ |
| Number of students | $40$ | $30$ | $20$ | $10$ |
Let $x_1, x_2, \dots, x_{100}$ be $100$ observations such that $\sum x_i = 0$,$\sum_{1 \le i < j \le 100} |x_i x_j| = 80000$,and the mean deviation from their mean is $5$. Then their standard deviation is:
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