We are given that there are 3 rotten apples and 7 good apples. 4 apples are drawn randomly without replacement. We need to find the value of \( 10(\mu^2 + \sigma^2) \), where \( \mu \) is the mean and \( \sigma^2 \) is the variance of the random variable \( X \), which represents the number of rotten apples drawn.
Step 1: Defining the Problem.
The number of rotten apples drawn, \( X \), follows a hypergeometric distribution, since we are drawing without replacement. The probability mass function (PMF) of \( X \) for a hypergeometric distribution is given by: \[ P(X = k) = \frac{\binom{3}{k} \binom{7}{4-k}}{\binom{10}{4}}, \] where: - \( 3 \) is the number of rotten apples,
- \( 7 \) is the number of good apples,
- \( 4 \) is the total number of apples drawn,
- \( k \) is the number of rotten apples drawn.
Step 2: Mean and Variance of Hypergeometric Distribution.
For a hypergeometric distribution, the mean \( \mu \) and variance \( \sigma^2 \) are given by the formulas: \[ \mu = \frac{nK}{N}, \quad \sigma^2 = \frac{nK(N-K)(N-n)}{N^2(N-1)}, \] where: - \( n = 4 \) (number of draws),
- \( K = 3 \) (total number of rotten apples),
- \( N = 10 \) (total number of apples).
Calculating the Mean \( \mu \): \[ \mu = \frac{4 \times 3}{10} = \frac{12}{10} = 1.2. \]
Calculating the Variance \( \sigma^2 \): \[ \sigma^2 = \frac{4 \times 3 \times (10 - 3) \times (10 - 4)}{10^2 \times (10 - 1)} = \frac{4 \times 3 \times 7 \times 6}{100 \times 9} = \frac{504}{900} = 0.56. \]
Step 3: Calculating \( 10(\mu^2 + \sigma^2) \). Now, calculate \( 10(\mu^2 + \sigma^2) \): \[ \mu^2 = (1.2)^2 = 1.44, \quad \sigma^2 = 0.56, \] \[ \mu^2 + \sigma^2 = 1.44 + 0.56 = 2. \] \[ 10(\mu^2 + \sigma^2) = 10 \times 2 = 20. \]
Thus, the value of \( 10(\mu^2 + \sigma^2) \) is \( 20 \), and the correct answer is option (4).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,