Q.
Consider the following frequency distribution
x
A
2A
3A
4A
5A
f
2
1
1
1
1
where, $A$ is a positive integer and has variance $160$. Then
the value of $A$ is.
| x | A | 2A | 3A | 4A | 5A |
| f | 2 | 1 | 1 | 1 | 1 |
Statistics
Solution:
$x_i$
$f_i$
$ f_i x_i$
$f_i x_i^2$
A
2
2A
$2A^2$
2A
1
2A
$4A^2$
3A
1
3A
$9A^2$
4A
1
4A
$16A^2$
5A
1
5A
$25A^2$
6A
1
6A
$36A^2$
Total
7
22A
$92A^2$
$\because \sigma^{2}=\frac{\sum f _{ i } x _{ i }^{2}}{\sum f _{ i }}-\left(\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}\right)$
$\Rightarrow 160=\frac{92 A ^{2}}{7}-\left(\frac{22 A }{7}\right)^{2}$
$\Rightarrow 160=\frac{92 A ^{2}}{7}-\frac{484 A ^{2}}{49} $
$\Rightarrow 160=\frac{92 \times 7 A ^{2}-484 A ^{2}}{49}$
$\Rightarrow 160 \times 49=644 A ^{2}-484 A ^{2} $
$\Rightarrow 160 A ^{2}=7840$
$\Rightarrow A ^{2}=\frac{7840}{160}$
$ \Rightarrow A ^{2}=49$
$ \Rightarrow A =\pm 7$
$A =7$ as $A$ is a positive integer.
| $x_i$ | $f_i$ | $ f_i x_i$ | $f_i x_i^2$ |
|---|---|---|---|
| A | 2 | 2A | $2A^2$ |
| 2A | 1 | 2A | $4A^2$ |
| 3A | 1 | 3A | $9A^2$ |
| 4A | 1 | 4A | $16A^2$ |
| 5A | 1 | 5A | $25A^2$ |
| 6A | 1 | 6A | $36A^2$ |
| Total | 7 | 22A | $92A^2$ |