If the mean of the data: 7,8,9,7,8,7,$\lambda$,8 is 8, then the variance of this data is :-
The dataset is given as: 7, 8, 9, 7, 8, 7, $\lambda$, 8. There are 8 observations (n=8).
The mean ($\mu$) is given as 8.
The formula for the mean is: $\mu = \frac{\sum x_i}{n}$
Substituting the values:
$8 = \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8}$
$8 = \frac{54 + \lambda}{8}$
Multiply both sides by 8:
$64 = 54 + \lambda$
Solve for $\lambda$:
$\lambda = 64 - 54 = 10$
Now the complete dataset is: 7, 8, 9, 7, 8, 7, 10, 8.
The mean ($\mu$) is 8.
The formula for variance ($\sigma^2$) is: $\sigma^2 = \frac{\sum (x_i - \mu)^2}{n}$
Calculate the squared differences from the mean:
Sum of the squared differences:
$\sum (x_i - \mu)^2 = 1 + 0 + 1 + 1 + 0 + 1 + 4 + 0 = 8$
Calculate the variance:
$\sigma^2 = \frac{8}{8} = 1$
The variance of the data is 1.
Let the mean and variance of 8 numbers $-10, -7, -1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively.
Then the mean of 4 numbers $x, y, x + y + 1, |x - y|$ is :
| $x_i$ | 5 | 7 | 9 | 10 | 12 | 15 |
| $f_i$ | 8 | 6 | 2 | 2 | 2 | 6 |
If the mean and median of the data
| x | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | |
| f | 3 | 6 | 2 | x | y | $\Sigma f = 20$ |
are equal, then $xy^2$ is equal to
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2\times2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :
Let the mean and variance of 8 numbers $-10, -7, -1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively.
Then the mean of 4 numbers $x, y, x + y + 1, |x - y|$ is :
| $x_i$ | 5 | 7 | 9 | 10 | 12 | 15 |
| $f_i$ | 8 | 6 | 2 | 2 | 2 | 6 |