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:

  • A
    $11$
  • B
    $9$
  • C
    $10$
  • D
    $12$

Explore More

Similar Questions

The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5, 7, 10, 12, 14, 15,$ then the absolute difference of the remaining two observations is

Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$ (where $\alpha < \beta$) be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is:

The mean of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. If an observation $x_{q}$ is replaced by $x_{q}^{\prime}$, then the new mean is:

The frequency distribution of daily working expenditure of families in a locality is as follows:
Expenditure in $Rs. (x)$ $0-50$ $50-100$ $100-150$ $150-200$ $200-250$
No. of families $(f)$ $24$ $33$ $37$ $b$ $25$

If the mode of the distribution is $Rs. 140$,then the value of $b$ is:

The mean of $5$ numbers is $18$. If one number is excluded,their mean becomes $16$. Then the excluded number is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo