Statement – II : The mean, median and mode do not coincide in the normal distribution.
Statement – III : The binomial distribution is symmetrical for any value of p (probability of success).
Codes :
We need to evaluate the truthfulness of three statements regarding probability distributions.
Statement I Analysis
The exponential distribution with rate parameter $\mu$ (often denoted as $\lambda$) has a probability density function (PDF) of the form $f(x; \mu) = \mu e^{-\mu x}$ for $x \ge 0$. The expected value (mean) and variance for this distribution are:
- Mean: $E[X] = \frac{1}{\mu}$
- Variance: $Var(X) = \frac{1}{\mu^2}$
Therefore, Statement – I is true.
Statement II Analysis
The normal distribution is a symmetrical probability distribution. In any symmetrical distribution, the mean, median, and mode are equal and coincide at the center of the distribution.
Thus, Statement – II, which claims they do not coincide, is false.
Statement III Analysis
The binomial distribution, denoted $B(n, p)$, represents the number of successes in a fixed number of independent Bernoulli trials. This distribution is symmetrical only when the probability of success, $p$, is exactly $0.5$. For any other value of $p$ ($p \ne 0.5$), the distribution is skewed.
Therefore, Statement – III, claiming symmetry for *any* value of $p$, is false.
Conclusion
Based on the analysis:
- Statement – I is true.
- Statement – II is false.
- Statement – III is false.
This corresponds to the option where Statement – I is true and Statements – II & III are false.
Important Questions from Probability Distribution
If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:
For the distribution with unknown θ
\(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)
We set the testing of hypothesis H 0 ∶ θ = 1 vs H 1 ∶ θ = 2. When the critical region X ≥ 0.4, the value of probability of type-II error is:
For the cumulative distribution function \(F(x) = \left\{ {\begin{array}{*{20}{c}} {0;x < - 1}\\ {\frac{1}{2}{{(x + 1)}^2}; - 1 \le x < 0}\\ {1 - \frac{{{{(1 - x)}^2}}}{2};0 \le x < 1}\\ {1.1 < x < \infty } \end{array}} \right.\)
the upper quartile point is
Let the joint probability density function of \( (X, Y) \) be
\[f(x, y) = \begin{cases} 6xy^2 & \text{if } 0 < x < 1, 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\]
Then \( P\left(\frac{1}{2} < X < \frac{3}{4}\right) \) is:
Let X and Y have the joint p.m.f. f(x, y) = x + y / 21, where x = 1, 2, 3 and y = 1, 2. The marginal p.m.f. of X is:
