AMU MCA Probability Distribution Previous Year Question Paper and Solution with PDF

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AMU MCA Previous Year Questions (PYQs)

AMU MCA Probability Distribution PYQ

Qus : 1 AMU MCA PYQ

If $X$ has an exponential distribution with mean $3$, then the variance of $X$ is

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Solution

For exponential distribution:

Mean $= \frac{1}{\lambda}$

Variance $= \frac{1}{\lambda^2}$

Given mean $=3 \Rightarrow \lambda=\frac{1}{3}$

Variance $=\frac{1}{(1/3)^2}=9$

Qus : 2 AMU MCA PYQ

Suppose random variable $X$ has possible values $1,2,3,\dots$ and $P(X=j)=\frac{1}{2^j}$, $j=1,2,3,\dots$ then $P(X \text{ is even})$ equals

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Solution

Even values: $2,4,6,\dots$

$P(\text{even})=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots$

$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots$

This is geometric series:

$=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}$

Qus : 3 AMU MCA PYQ

If $X$ has discrete uniform distribution with pmf $f(x)=\frac{1}{8}$ for $x=1,2,\dots,8$, then the mean of the distribution is

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Solution

Mean of uniform integers $1$ to $8$:

$=\frac{1+8}{2}=4.5$

Qus : 4 AMU MCA PYQ

If $P(X=0)=1-P(X=1)$ and $E(X)=3V(X)$, the value of $P(X=0)$ is

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Solution

Let $p=P(X=1)$

Then $P(X=0)=1-p$

$E(X)=p$

$V(X)=p(1-p)$

Given $p=3p(1-p)$

$1=3(1-p)$

$1=3-3p$

$3p=2$

$p=\frac{2}{3}$

Therefore $P(X=0)=1-p=\frac{1}{3}$

Qus : 5 AMU MCA PYQ

If $X$ is a discrete random variable taking positive integer values and possesses memory-less property, then the distribution of $X$

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Solution

The only discrete distribution having memory-less property is the Geometric distribution.

Qus : 6 AMU MCA PYQ

If $X \sim N(\mu,\sigma^2)$, then $Z^2=\left(\frac{X-\mu}{\sigma}\right)^2$ is a Chi-square variate with

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Solution

$Z=\frac{X-\mu}{\sigma} \sim N(0,1)$

Square of standard normal variable follows $\chi^2$ distribution with 1 d.f.

Qus : 7 AMU MCA PYQ

The points of inflexion of a normal distribution with mean $\mu$ and variance $\sigma^2$ are

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Solution

For normal curve, points of inflexion are at

$\mu \pm \sigma$

Qus : 8 AMU MCA PYQ

Let $X_1, X_2, \dots, X_n$ be independently and identically distributed $N(\mu,\sigma^2)$. Which of the following is simple hypothesis under $H_0$?

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Solution

A simple hypothesis completely specifies parameter(s).

Thus $\mu=\mu_0$ with $\sigma^2$ known is simple.

Qus : 9 AMU MCA PYQ

If the random variable $X$ has the density function $f(x)=e^{x-2}$ for $x<2$, and $f(x)=0$ otherwise, then the 75th percentile and third quartile is

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Solution

Qus : 10 AMU MCA PYQ

A discrete random variable $X$ satisfies the memory less property, then the random variable $X$ follows

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Solution

Qus : 11 AMU MCA PYQ

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

Qus : 12 AMU MCA PYQ

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

Qus : 13 AMU MCA PYQ

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 $= npq = 3 \times \frac12 \times \frac12 = \frac{3}{4}$

Qus : 14 AMU MCA PYQ

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 $= \lambda = 3$ $E(X-6)^2 = \text{Var}(X) + (E(X) - 6)^2$ $= 3 + (3 - 6)^2 = 3 + 9 = 12$

Qus : 15 AMU MCA PYQ

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.

Qus : 16 AMU MCA PYQ

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.

Qus : 17 AMU MCA PYQ

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 $\frac12$.

Qus : 18 AMU MCA PYQ

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 $\lambda = 2$. $P(X \le 1) = P(0)+P(1)$ $= e^{-2} + 2e^{-2} = 3e^{-2}$

Qus : 19 AMU MCA PYQ

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, $= [F(x)]^n = (1-e^{-x/\theta})^n$

Qus : 20 AMU MCA PYQ

A box contains tickets numbered $1$ to $N$. Let $X$ be the largest number drawn in $n$ random drawings with replacement. Then the probability $P(X = k)$ is:

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Solution

$P(X \le k) = (k/N)^n$

$P(X \le k-1) = ((k-1)/N)^n$

So,

$P(X = k) = P(X \le k) - P(X \le k-1)$

$= (k/N)^n - ((k-1)/N)^n$

Qus : 21 AMU MCA PYQ

Let $x_1=2.2,\ x_2=4.1,\ x_3=3.4,\ x_4=4.5,\ x_5=1.1,\ x_6=5.7$ be sample from $U(0,\theta)$. Then MLE of $\theta$ is:

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Solution

For $U(0,\theta)$, $\hat{\theta}_{MLE}=\max(x_i)$ Maximum = $5.7$

Qus : 22 AMU MCA PYQ

If $X_1,\dots,X_n$ are Bernoulli with probability $p$, then constant estimate of $p(1-p)$ is

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Solution

Since $\hat{p}=\bar{X}$, Estimate of $p(1-p)$ is $\bar{X}(1-\bar{X})$

Qus : 23 AMU MCA PYQ

Let $E(X)=\mu$, $Var(X)=9$. Smallest $m$ such that $P(|X-\mu|
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Solution

Using Chebyshev inequality:

$P(|X-\mu|\ge m)\le \frac{Var(X)}{m^2}$

$1-\frac{9}{m^2}\ge 0.99$

$\frac{9}{m^2}\le 0.01$

$m^2\ge 900$

$m=30$

Qus : 24 AMU MCA PYQ

If the mean and variance of a binomial variate $X$ are 8 and 4 respectively, then $P(X<3)$ equals:

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Solution

For binomial distribution:

Mean $=np=8$

Variance $=np(1-p)=4$

So,

$8(1-p)=4$

$1-p=\frac12$

$p=\frac12$

Then $n=16$

Now,

$P(X<3)=P(0)+P(1)+P(2)$

$= \frac{\binom{16}{0}+\binom{16}{1}+\binom{16}{2}}{2^{16}}$

$= \frac{1+16+120}{2^{16}}$

$= \frac{137}{2^{16}}$

Qus : 25 AMU MCA PYQ

If $X$ is the number of heads obtained in four tosses of a balanced coin. Define $ Y=\dfrac{1}{1+X} $ Find $P\left(Y=\frac{1}{3}\right)$

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Solution

$Y=\frac{1}{3} \Rightarrow 1+X=3 \Rightarrow X=2$ So we need: $P(X=2)$ For 4 tosses: $P(X=2)=\binom{4}{2}\left(\frac12\right)^4$ $=\frac{6}{16}=\frac{3}{8}$

Qus : 26 AMU MCA PYQ

If $g$ is convex and $X$ is random variable, then:

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Solution

By Jensen's inequality: $E[g(X)]\ge g(E[X])$

Qus : 27 AMU MCA PYQ

If $X$ follows geometric distribution, then: $P(X\ge j+k|X\ge j)=$

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Solution

Geometric distribution has memoryless property: $P(X\ge j+k|X\ge j)=P(X\ge k)$

Qus : 28 AMU MCA PYQ

Let $X\sim N(\mu,\sigma^2)$ where both unknown. Test hypothesis:

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Solution

Simple hypothesis specifies both parameters. So correct simple hypothesis: $H_0:\mu=2,\sigma=4$

Qus : 29 AMU MCA PYQ

The degree of peakedness is called:

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Solution

The statistical measure that describes the peakedness of a distribution is called Kurtosis.

Qus : 30 AMU MCA PYQ

The skewness in a binomial distribution will be zero, if

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Solution

Skewness of binomial:

$\gamma = \dfrac{1-2p}{\sqrt{np(1-p)}}$

Zero when:

$1-2p=0$

$p=\frac12$

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