The mean of $100$ items is $49$. It was discovered that three items which should have been $60, 70, 80$ were wrongly read as $40, 20, 50$ respectively. The correct mean is

  • A
    $48$
  • B
    $82\frac{1}{2}$
  • C
    $50$
  • D
    $80$

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Similar Questions

The average income of individuals in one group is $\bar{x}$ and the average income of another group is $\bar{y}$. If the ratio of the number of individuals in both groups is $4:3$,what is the average income of the combined group?

The number of values of $a \in N$ such that the variance of $3, 7, 12, a, 43-a$ is a natural number is (Mean $= 13$).

The mean and variance of seven observations are $8$ and $16$ respectively. If $5$ of the observations are $2, 4, 10, 12, 14$, then the square root of the product of the remaining two observations is (in $\sqrt{3}$)

$A$ market with $3900$ operating firms has the following distribution for firms arranged according to various income groups of workers. If a histogram for the above distribution is constructed,the highest bar in the histogram would correspond to the class:
Income group No. of firms
$150-300$ $300$
$300-500$ $500$
$500-800$ $900$
$800-1200$ $1000$
$1200-1800$ $1200$

Consider the following frequency distribution:
Value $4$ $5$ $8$ $9$ $6$ $12$ $11$
Frequency $5$ $f_1$ $f_2$ $2$ $1$ $1$ $3$

Suppose that the sum of the frequencies is $19$ and the median of this frequency distribution is $6$. For the given frequency distribution,let $\alpha$ denote the mean deviation about the mean,$\beta$ denote the mean deviation about the median,and $\sigma^2$ denote the variance. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.
List-$I$ List-$II$
$(P) \ 7f_1+9f_2$ is equal to $(1) \ 146$
$(Q) \ 19\alpha$ is equal to $(2) \ 47$
$(R) \ 19\beta$ is equal to $(3) \ 48$
$(S) \ 19\sigma^2$ is equal to $(4) \ 145$
$(5) \ 55$

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