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The mean of the numbers \( a, b, 8, 5, 10 \) is \( 6 \) and their variance is \( 6.8 \). If \( M \) is the mean deviation of the numbers about the mean, then \( 25M \) is equal to:

The sum and product of the mean and variance of a binomial distribution are \( 82\cdot 5 \) and \( 1350 \) respectively. Then the number of trials in the binomial distribution is __________.

The mean and standard deviation of \( 15 \) observations are found to be \( 8 \) and \( 3 \) respectively. On rechecking it was found that, in the observations, \( 20 \) was misread as \( 5 \). Then, the correct variance is equal to ____________.

The mean and variance of \( 10 \) observations were calculated as \( 15 \) and \( 15 \) respectively by a student who took by mistake \( 25 \) instead of \( 15 \) for one observation. Then, the correct standard deviation is ______.

If the mean deviation about median for the numbers \( 3, 5, 7, 2k, 12, 16, 21, 24 \), arranged in the ascending order, is \( 6 \) then the median is:

The mean and variance of a binomial distribution are \( \alpha \) and \( \dfrac{\alpha}{3} \) respectively. If \( P(X=1)=\dfrac{4}{243} \), then \( P(X=4 \text{ or } 5) \) is equal to:

If the mean deviation about the mean of the numbers \( 1, 2, 3, \ldots, n \), where \( n \) is odd, is \( \frac{5(n+1)}{n} \), then \( n \) is equal to __________.

Suppose a class has \( 7 \) students. The average marks of these students in the mathematics examination is \( 62 \), and their variance is \( 20 \). A student fails in the examination if he/she gets less than \( 50 \) marks, then in worst case, the number of students can fail is \( \_ \_ \_ \_ \_ \_ \).

Let the mean and the variance of \( 20 \) observations \( x_1,x_2,\ldots,x_{20} \) be \( 15 \) and \( 9 \), respectively. For \( \alpha \in R \), if the mean of \( (x_1+\alpha)^2,(x_2+\alpha)^2,\ldots,(x_{20}+\alpha)^2 \) is \( 178 \), then the square of maximum value of \( \alpha \) is equal to _________.

Let the mean and the variance of \( 5 \) observations \( x_1, x_2, x_3, x_4, x_5 \) be \( \frac{24}{5} \) and \( \frac{194}{25} \) respectively. If the mean and variance of the first \( 4 \) observation are \( \frac{7}{2} \) and \( a \) respectively, then \( (4a + x_5) \) is equal to:

The mean and standard deviation of \( 50 \) observations are \( 15 \) and \( 2 \) respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is \( 70 \). If the correct mean is \( 16 \), then the correct variance is equal to:

The number of values of \( a \in \mathbb{N} \) such that the variance of \( 3,7,12,a,43-a \) is a natural number is:

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JEE Main 2022 29th June Evening Shift

The number of values of \( a \in \mathbb{N} \) such that the variance of \( 3,7,12,a,43-a \) is a natural number is: