The mean of $10$ observations is $52$. If $12$ is subtracted from each observation,then the mean of the new observations so obtained is:

  • A
    $12$
  • B
    $26$
  • C
    $13$
  • D
    $40$

Explore More

Similar Questions

Draw a histogram for the following frequency distribution with unequal class length:
Class $0-10$ $10-20$ $20-40$ $40-70$ $70-100$
Frequency $15$ $13$ $28$ $36$ $27$

$A$ football player scored the following number of goals in $10$ matches:
$1, 3, 2, 5, 8, 6, 1, 4, 7, 9$
Since the number of matches is $10$ (an even number),the median is calculated as:
Median $= \frac{5^{\text{th}} \text{ observation} + 6^{\text{th}} \text{ observation}}{2}$
$= \frac{8 + 6}{2} = 7$
Is this the correct answer? If not,why?

If $\bar{x}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n}$ and $\bar{y}$ is the mean of $y_{1}, y_{2}, \ldots, y_{n}$. If $\bar{z}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n}, y_{1}, y_{2}, \ldots, y_{n}$,then $\bar{z}$ is equal to

The expenditure of a family on different heads in a month is given below:
Head Expenditure (in $Rs$)
Food $4000$
Education $2500$
Clothing $1000$
House Rent $3500$
Others $2500$
Savings $1500$

Draw a bar graph to represent the data above.

Heights (in $cm$) of $30$ girls of Class $IX$ are given below:
$140, 140, 160, 139, 153, 153, 146, 150, 148, 150, 152$
$146, 154, 150, 160, 148, 150, 148, 140, 148, 153, 138$
$152, 150, 148, 138, 152, 140, 146, 148$
Prepare a frequency distribution table for this data.

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