AMU MCA Probability Distribution Previous Year Questions (PYQs) – Page 2 of 4
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Showing 31 questions
While testing the independence of attributes using Chi-square distribution, suppose attribute $A$ is specified into three classes and attribute $B$ is classified into four classes, then the degree of freedom of Chi-square test is:
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Solution
f a random variable $X$ takes the values $1, 2, 3, 4$ such that
$2P(X=1)=3P(X=2)=P(X=3)=5P(X=4)$,
then $P(X=2)$ is:
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Solution
If $X$ follows a binomial distribution with parameters $n=100,\ p=\frac{1}{3}$, then $P(X=r)$ is maximum, when $r$ is equal to
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Solution
Mode of binomial:
$ r=\lfloor (n+1)p \rfloor $
$ =(101)\cdot\frac{1}{3}=33.67 $
$ \Rightarrow r=33 $
Three identical fair dice are thrown independently. Let $Y$ denote the number of dice showing even numbers on their upper faces. Then the variance of random variable $Y$ is:
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Solution
Each die has probability $p = \frac{1}{2}$ of showing an even number.
So $Y \sim \text{Binomial}(3, \frac12)$
Variance
$= npq = 3 \times \frac12 \times \frac12 = \frac{3}{4}$
A discrete random variable follows Poisson distribution with parameter $3$. The value of $E(X-6)^2$ is:
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Solution
For Poisson distribution,
Mean $= \lambda = 3$
Variance $= \lambda = 3$
$E(X-6)^2 = \text{Var}(X) + (E(X) - 6)^2$
$= 3 + (3 - 6)^2 = 3 + 9 = 12$
Let $X$ be a random variable having distribution function $F(x)$. Then which of the following statements may not be true?
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Solution
A distribution function is always right continuous, but it need not be left continuous.
Which of the following statements is not true?
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Solution
For geometric distribution, variance is $\frac{q}{p^2}$, which is not equal to 1 for mean 2.
Let $X$ be a continuous random variable such that $E|X| < \infty$ and
$P\left(X \ge \frac12 + x\right) = P\left(X \le \frac12 - x\right)$ for all $x \in \mathbb{R}$.
Then:
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Solution
The given condition shows symmetry about $\frac12$.
Hence both mean and median are $\frac12$.
A discrete random variable $X$ taking non-negative values has the following moment generating function
$M_X(t) = e^{2(e^t-1)},; -\infty < t < \infty$
Then, the value of $P(X \le 1)$ is:
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Solution
Given MGF corresponds to Poisson distribution with $\lambda = 2$.
$P(X \le 1) = P(0)+P(1)$
$= e^{-2} + 2e^{-2} = 3e^{-2}$
Let $X_1, X_2, … , X_n$ be i.i.d. random variables having pdf
$f(x) = (1/θ) e^{-x/θ},; 0 < x < ∞,; θ > 0$
The cdf of the largest order statistic
$X_{(n)} = max(X_1, X_2, … , X_n)$
is:
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Solution
First find the cdf of a single variable $X$:
$F(x)=P(X\le x)=\int_0^x \frac{1}{\theta}e^{-t/\theta},dt
=1-e^{-x/\theta}$
For the largest order statistic,
$F_{X_{(n)}}(x)=P(X_{(n)}\le x)$
Largest $\le x$ means all observations $\le x$:
$P(X_{(n)}\le x)=P(X_1\le x,\dots,X_n\le x)$
Since variables are independent,
$= [F(x)]^n = (1-e^{-x/\theta})^n$
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