Statistics

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JEE Main 2025 (Online) 4th April Evening ShiftQuestion 6

Let the mean and the standard deviation of the observation $2,3,3,4,5,7,a,b$ be $4$ and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is :

Statistics

$2$$\frac{1}{2}$$\frac{3}{4}$$1$

JEE Main 2021 (Online) 26th August Morning ShiftQuestion 7

The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35$. If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha,\beta)$ is :

Statistics

$(11,25)$$(11,26)$$(10.5,26)$$(10.5,25)$

JEE Main 2025 (Online) 3rd April Evening ShiftQuestion 8

Let the Mean and Variance of five observations $x_1=1$, $x_2=3$, $x_3=a$, $x_4=7$ and $x_5=b$, $a>b$, be $5$ and $10$ respectively. Then the Variance of the observations $n+x_n$, $n=1,2,\ldots,5$ is

Statistics

$16$$16.4$$17$$17.4$

JEE Main 2024 (Online) 1st February Evening ShiftQuestion 9

Consider $10$ observations $x_1,x_2,\ldots,x_{10}$ such that $\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\sum_{i=1}^{10}(x_i-\beta)^{2}=40$, where $\alpha,\beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{\beta}{\alpha}$ is equal to :

Statistics

$1$$2$$\frac{3}{2}$$\frac{5}{2}$

JEE Main 2023 (Online) 15th April Morning ShiftQuestion 10

The mean and standard deviation of $10$ observations are $20$ and $8$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is

Statistics

$11$$12$$13$$14$