1
MHT CET 2025 26th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

A random variable $X$ has the following probability distribution :

$$ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0.1 & 0.2 & 0.3 & 0.4 \\ \hline \end{array} $$

The mean and standard deviation of $X$ are respectively

A
2 and 3
B
3 and 1
C
3 and $\sqrt{2}$
D
2 and 1
2
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The probability distribution of a random variable X is given by

$$ \begin{array}{|l|c|c|c|c|c|} \hline \mathrm{X}=x_i & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}\left(\mathrm{X}=x_i\right) & 0.4 & 0.3 & 0.1 & 0.1 & 0.1 \\ \hline \end{array} $$

Then the variance of X is

A
1.76
B
2.45
C
3.2
D
4.8
3
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let mean and standard deviation of probability distribution

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X}=x & -3 & 0 & 1 & \alpha \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{4} & \mathrm{~K} & \frac{1}{4} & \frac{1}{3} \\ \hline \end{array} $$

be $\mu$ and $\sigma$ respectively and if $\sigma-\mu=2$ then $\sigma=$

A
$\frac{3}{2}$
B
$\frac{5}{2}$
C
$\frac{7}{2}$
D

$\frac{9}{2}$

4
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The mean and variance of seven observations are 8 and 16 respectively. If five of the observations are $2,4,10,12,14$, then the product of remaining two observations is

A
45
B
44
C
48
D
40
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