The median of the normal distribution with mean and variance \(\mu\) and \(\sigma^2\) is:
\(\mu\)
For a normal distribution, the median equals the mean \(\mu\) due to symmetry. The median is the central value where 50% of the data lies on either side, coinciding with \(\mu\) in symmetric distributions like the normal curve.
Similar Questions
Out of 100 numbers, 20 are 4's, 40 are 5's, 30 are 6's, and the remaining are 7's. The arithmetic mean of the numbers is:
Let a continuous random variable \( X \) have probability density function (pdf):
\[f(x) = \begin{cases} -0.75 \, x^2 + 1.5x & \text{for } 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}\]
Find the mode of \( X \).
If the arithmetic mean is 25 and geometric mean is 15, then the value of the Harmonic mean is equal to:
Important Questions from Mean and Variance of Random variables
X is a non-negative integer valued random variable with
\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)
Then, mean and variance of X are respectively
Out of 100 numbers, 20 are 4's, 40 are 5's, 30 are 6's, and the remaining are 7's. The arithmetic mean of the numbers is:
Let a continuous random variable \( X \) have probability density function (pdf):
\[f(x) = \begin{cases} -0.75 \, x^2 + 1.5x & \text{for } 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}\]
Find the mode of \( X \).
If the arithmetic mean is 25 and geometric mean is 15, then the value of the Harmonic mean is equal to:
A random variable X has the distribution law as given below:
X
1
2
3
P(X = x)
0.3
0.4
0.3
The variance of the distribution is:
