Random Variables & Distribution Functions MCQ Quiz - Objective Question with Answer for Random Variables & Distribution Functions - Download Free PDF

Solution:  

we use the properties of the cumulative distribution function (CDF).

Step 1: Calculate $P(1<X\le 2)$

Using the CDF properties, for $a<X\le b$ :

$P(a<X\le b)=F(b)-F(a)$.

Here, a = 1 and b = 2 .

For 0 ≤ x < 3 , the CDF is given by $F(x)=\frac{x+2}{5}$ .

Substituting these values:

$F(2)=\frac{2+2}{5}=\frac{4}{5},F(1)=\frac{1+2}{5}=\frac{3}{5}.$

Thus:

$P(1<X\le 2)=F(2)-F(1)=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}.$

Step 2: Calculate P(X = 0)

The probability P(X = 0) corresponds to the probability mass at X = 0 .

Since the given random variable is continuous, P(X = 0) = 0 .

Step 3: Add the probabilities

Combining the results:

$P(1<X\le 2)+P(X=0)=\frac{1}{5}+0=\frac{1}{5}.$

Hence the correct option is (1)

Concept:

1. Law of Large Numbers (LLN):

The Law of Large Numbers states that as the sample size n increases, the sample average (or sum) of i.i.d.

random variables converges to the expected value of the variable. For example, in Option 3, $\frac{1}{n}\sum _{i=1}^n{X}_i^2$ 

converges to the expected value $E(X_1^2)=1$, because $X_1$ has variance 1.

2. Central Limit Theorem (CLT):

The Central Limit Theorem tells us that the sum (or scaled average) of i.i.d. random variables with finite mean and

variance converges in distribution to a normal distribution as $n\to \mathrm{\infty }$ . For instance, in Option 4,  $\frac{\sum _{i=1}^n{X}_i}{\sqrt{n}}$ behaves like a

standard normal random variable as  $n\to \mathrm{\infty }$ , converging to N(0, 1) .

3. Probability Limits:

For certain random variables, their distribution converges to a fixed probability value. In Options 1 and 4,

the probability of the standardized sum being less than or equal to 0 converges to $\frac{1}{2}$, which is the probability

that a standard normal variable is less than or equal to 0.

Explanation:

Option 1: This expression involves the ratio of two terms: $\sqrt{n}\sum _{i=1}^n{X}_{i}and\sum _{i=1}^n{X}_i^2$.

The numerator,$\sqrt{n}\sum _{i=1}^n{X}_i$, grows like $O(\sqrt{n})$ by the Central Limit Theorem since the sum of i.i.d. random

variables with mean zero and variance 1 tends to have a normal distribution.

The denominator, $\sum _{i=1}^n{X}_i^2$ , grows like O(n) because it is the sum of the squares of i.i.d. random variables with variance 1.

As $n\to \mathrm{\infty }$, the ratio $\frac{\sqrt{n}\sum _{i=1}^n{X}_i}{\sum _{i=1}^n{X}_i^2}$ tends to zero, so the probability converges to the probability of the random variable being

less than or equal to 0. Hence, Option 1 is true, and the limit will be 1/2 because this becomes a standard normal random variable under large n .

Option 2: The numerator $\sum _{i=1}^n{X}_i$ behaves like $O(\sqrt{n})$ , and the denominator $\sum _{i=1}^n{X}_i^2$ behaves like O(n) . Therefore, the ratio $\frac{\sum _{i=1}^n{X}_i}{\sum _{i=1}^n{X}_i^2}$ behaves

like $O(1/\sqrt{n})$ and converges to 0 as $n\to \mathrm{\infty }$ . Thus, Option 2 is true.

Option 3: By the Law of Large Numbers (LLN), the average of the squares of i.i.d. random variables converges to the

expected value of $X_1^2$, which is 1, as $n\to \mathrm{\infty }$. Thus, Option 3 is true.

Option 4: 
$$
By the Central Limit Theorem (CLT), $\frac{\sum _{i=1}^n{X}_i}{\sqrt{n}}$ converges in distribution to a standard normal random variable N(0, 1) .

The probability that a standard normal variable is less than or equal to 0 is $P(Z\le 0)=\frac{1}{2}$. Hence, Option 4 is true.
 

All four options are correct.

Concept:

The sample mean gives an average of the data points in a sample, and the sample variance

measures the spread of the data.

The normal distribution is a continuous probability distribution that is symmetric around the mean,

with most values clustering around the center.

Explanation:

Option 1: The mean of $\overline{X}$ is 2 and the mean of $\overline{Y}$ is -2 . Thus, the mean of $\overline{X}+\overline{Y}$ is
 

$E[\overline{X}+\overline{Y}]=2+(-2)=0$

The variances add for independent random variables:

  $\text{Var}(\overline{X})=\frac{{\sigma }^2}{n}=\frac{4}{12}=\frac{1}{3},\text{Var}(\overline{Y})=\frac{5}{15}=\frac{1}{3}$

Therefore, the variance of $\overline{X}+\overline{Y}$ is $\text{Var}(\overline{X}+\overline{Y})=\text{Var}(\overline{X})+\text{Var}(\overline{Y})=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

Thus, $\overline{X}+\overline{Y}$ is distributed as $N(0,\frac{2}{3})$ . This statement is true.

Option 2: $S_1^2$ and $S_2^2$  are independent chi-squared random variables scaled by their respective degrees of freedom.

The degrees of freedom for $S_1^2$ is 11 and for  $S_2^2$ is 14 .

The weighted sum does not follow a chi-squared distribution directly. The combined degrees of

freedom would be 11 + 14 = 25 , not 26 . This statement is false.

Option 3: The ratio $\frac{S_1^2}{S_2^2}$ is distributed as $F_{11,14}$ because it represents the ratio of two independent chi-squared

variables divided by their respective degrees of freedom. Multiplying by a constant (in this case, $\frac{5}{4}$)

does not change the F-distribution's form, just scales it. This statement is true.

Option 4: The mean of $\overline{Y}$ is -2 . The distribution of  $\overline{Y}$ is  $N(-2,\frac{5}{15})=N(-2,\frac{1}{3})$ . 

The quantity  $\overline{Y}$ - 2 shifts the mean to -4 . Scaling a normal distribution does not yield a t-distribution.

Instead,  $2\sqrt{3}(\overline{Y}-2)$ results in a normal distribution, not a t-distribution. This statement is false.

Hence, option 1) and 3) are correct.

Concepts Used:

1. Cumulative Distribution Function (CDF):

The CDF F(x) gives the probability that the random variable X takes a value less than or equal to x . That is, F(x) = P(X ≤ x) .

2. Finding Probability Using the CDF:

The probability that the random variable X lies within a certain interval (a, b] is given by:
     
 P(a < X ≤ b) = F(b) - F(a)

3. Probability at a Specific Point (Jump Discontinuity):

The probability at a specific point x = c is the difference in the CDF just to the right and just to the left of c :
     
P(X = c) = F(c⁺) - F(c^-)

Explanation -

We are given a cumulative distribution function (CDF) F(x) of a random variable X as:

F(x) = $\{\begin{array}{ll}0,& \text{if }x<0\\ \frac{x+1}{3},& \text{if }0\le x<1\\ 1,& \text{if }x\ge 1\end{array}$

From the definition of the probability from the CDF:  P(a < X $\le$ b) = F(b) - F(a)

In this case, we need to calculate $F\left(\frac{3}{4}\right)andF\left(\frac{1}{3}\right)$.

Since $0\le \frac{1}{3}<1$  and $0\le \frac{3}{4}<1$ , we use the formula $F(x)=\frac{x+1}{3}$ for both 1/3  and 3/4 :

$F\left(\frac{1}{3}\right)=\frac{\frac{1}{3}+1}{3}=\frac{\frac{4}{3}}{3}=\frac{4}{9}$

$\Rightarrow F\left(\frac{3}{4}\right)=\frac{\frac{3}{4}+1}{3}=\frac{\frac{7}{4}}{3}=\frac{7}{12}$

Thus, the probability $P(\frac{1}{3}<X\le \frac{3}{4})$ is:

$P(\frac{1}{3}<X\le \frac{3}{4})=F\left(\frac{3}{4}\right)-F\left(\frac{1}{3}\right)=\frac{7}{12}-\frac{4}{9}$

= $\frac{21}{36}-\frac{16}{36}=\frac{5}{36}$

The probability at a point is the jump in the CDF at that point. We need to calculate F(0⁺) - F(0^-) .

From the CDF definition: F(0⁺) = F(0) =$\frac{0+1}{3}=\frac{1}{3}$

⇒ F(0^-) = 0

Thus, $P(X=0)=F(0^{+})-F(0^{-})=\frac{1}{3}-0=\frac{1}{3}=\frac{12}{36}$

Now, we add the two results:

$P(\frac{1}{3}<X\le \frac{3}{4})+P(X=0)=\frac{5}{36}+\frac{12}{36}=\frac{17}{36}$

Thus, the final answer is 17/36.

Concept:

Let F(x) be a cumulative distribution function then

(i) $\underset{x\to -\mathrm{\infty }}{lim}F(x)$ = 0, $\underset{x\to \mathrm{\infty }}{lim}F(x)$ = 1

(ii) F is non-decreasing function

Explanation:

(2): F(x) = $\{\begin{array}{ll}\frac{1}{2+{\mathrm{x}}^2}& \text{ if }\mathrm{x}<0,\\ \frac{2+{\mathrm{x}}^2}{3+2{\mathrm{x}}^2}& \text{ if }\mathrm{x}\ge 0\end{array}$

$\underset{x\to \mathrm{\infty }}{lim}F(x)$ = $\underset{x\to \mathrm{\infty }}{lim}\frac{2+x^2}{3+2x^2}$ = $\underset{x\to \mathrm{\infty }}{lim}\frac{2x}{4x}$ = $\frac{1}{2}$ (Using L'hospital rule), not satisfying

Option (2) is false

(3): F(x) = $\{\begin{array}{ll}\frac{1}{2+{\mathrm{x}}^2}& \text{ if }\mathrm{x}<0\text{, }\\ \frac{2\cos (\mathrm{x})+{\mathrm{x}}^2}{4+{\mathrm{x}}^2}& \text{ if }\mathrm{x}\ge 0\end{array}$

$\underset{x\to \mathrm{\infty }}{lim}F(x)$ = $\underset{x\to \mathrm{\infty }}{lim}\frac{2\cos x+x^2}{4+x^2}$ 

 = $\underset{x\to \mathrm{\infty }}{lim}\frac{-2\sin x+2x}{2x}$ (Using L'hospital rule)

= $\underset{x\to \mathrm{\infty }}{lim}(-\frac{\sin x}{x}+1)$ - 1 + 1 = 0, not satisfying

Option (3) is false

(4): F(x) = $\{\begin{array}{ll}\frac{1}{2+{\mathrm{x}}^2}& \text{ if }\mathrm{x}<0,\\ \frac{1+{\mathrm{x}}^2}{4+{\mathrm{x}}^2}& \text{ if }\mathrm{x}\ge 0\end{array}$

f(0^-) = 1/2 and f(0⁺) = 1/4 and $\frac{1}{2}>\frac{1}{4}$ so F(x) not satisfying the property "F is non-decreasing function"

Option (4) is false

Therefore option (1) is correct

Concept:

A parametric function f(λ) is said to be estimable if there exist g(X) such that E(g(X)) = f(λ) otherwise it is called not estimable.

Explanation:

Given X be a Poisson random variable with mean λ

So E(X) = λ and Var(X) = λ

We have to find the parametric function which is not estimable.

(2): E(X) = λ so here we are getting a function g(X) = X

So it is estimable

Option (2) is false

(3): E(X²) = [E(X)]² + Var(X) = λ² + λ

So E(X² - X) = E(X2) - E(X) =  λ2 + λ - λ = λ2 

Here we are getting the function g(X) = X2 - X

 So it is estimable

Option (3) is false

(4): E$(\sum \frac{(-1{)}^x{\lambda }^x}{x!})$ = e−λ​

Here we are getting the function g(X) = $(\sum \frac{(-1{)}^x{\lambda }^x}{x!})$

 So it is estimable 

Option (4) is false

Hence option (1) is correct

Concepts Used:

1. Cumulative Distribution Function (CDF):

The CDF F(x) gives the probability that the random variable X takes a value less than or equal to x . That is, F(x) = P(X ≤ x) .

2. Finding Probability Using the CDF:

The probability that the random variable X lies within a certain interval (a, b] is given by:
     
 P(a < X ≤ b) = F(b) - F(a)

3. Probability at a Specific Point (Jump Discontinuity):

The probability at a specific point x = c is the difference in the CDF just to the right and just to the left of c :
     
P(X = c) = F(c⁺) - F(c^-)

Explanation -

We are given a cumulative distribution function (CDF) F(x) of a random variable X as:

F(x) = $\{\begin{array}{ll}0,& \text{if }x<0\\ \frac{x+1}{3},& \text{if }0\le x<1\\ 1,& \text{if }x\ge 1\end{array}$

From the definition of the probability from the CDF:  P(a < X $\le$ b) = F(b) - F(a)

In this case, we need to calculate $F\left(\frac{3}{4}\right)andF\left(\frac{1}{3}\right)$.

Since $0\le \frac{1}{3}<1$  and $0\le \frac{3}{4}<1$ , we use the formula $F(x)=\frac{x+1}{3}$ for both 1/3  and 3/4 :

$F\left(\frac{1}{3}\right)=\frac{\frac{1}{3}+1}{3}=\frac{\frac{4}{3}}{3}=\frac{4}{9}$

$\Rightarrow F\left(\frac{3}{4}\right)=\frac{\frac{3}{4}+1}{3}=\frac{\frac{7}{4}}{3}=\frac{7}{12}$

Thus, the probability $P(\frac{1}{3}<X\le \frac{3}{4})$ is:

$P(\frac{1}{3}<X\le \frac{3}{4})=F\left(\frac{3}{4}\right)-F\left(\frac{1}{3}\right)=\frac{7}{12}-\frac{4}{9}$

= $\frac{21}{36}-\frac{16}{36}=\frac{5}{36}$

The probability at a point is the jump in the CDF at that point. We need to calculate F(0⁺) - F(0^-) .

From the CDF definition: F(0⁺) = F(0) =$\frac{0+1}{3}=\frac{1}{3}$

⇒ F(0^-) = 0

Thus, $P(X=0)=F(0^{+})-F(0^{-})=\frac{1}{3}-0=\frac{1}{3}=\frac{12}{36}$

Now, we add the two results:

$P(\frac{1}{3}<X\le \frac{3}{4})+P(X=0)=\frac{5}{36}+\frac{12}{36}=\frac{17}{36}$

Thus, the final answer is 17/36.

Given:-

X1, X2, ... are i.i.d. random variables having a χ2-distribution with 5 degrees of freedom.

Concept Used:-

The limiting distribution of the given expression can be found using the central limit theorem.

The central limit theorem states that the sum of many independent and identically distributed random variables, properly normalized, converges in distribution to a normal distribution.

Explanation:-

Here, we have n i.i.d. random variables with a χ²-distribution with 5 degrees of freedom.

The mean of each χ²-distributed variable is 5 and the variance is,

2 × 5 = 10

Therefore, the mean of the sum of n such variables is n5, and the variance is,

⇒ variance = (n × 10)

We can normalize the expression by subtracting the mean and dividing by the standard deviation. That is,

$\Rightarrow a[{\displaystyle \frac{(X_1+X_2+...+X_n-5n)}{\sqrt{n\times 10}}}]$

$=({\displaystyle \frac{a}{\sqrt{10}}})[{\displaystyle \frac{(X_1+X_2+...+X_n-5n)}{n}}]\sqrt{n}$

The term in the brackets on the right-hand side is the sum of n i.i.d. random variables with a mean of 0 and a variance of 1/2.

Therefore, by the CLT, this term converges in distribution to a standard normal distribution as n goes to infinity.

The overall expression converges in distribution to a normal distribution with mean zero and variance a²/10.

So, the limiting distribution of $a\left(\frac{X_1+\cdots +X_n-5n}{\sqrt{n}}\right)$ is ​​the standard normal distribution for an appropriate value of a.

Hence, the correct option is 3.

The correct answer is option 4 

we will update the solution as soon as possible.

The correct answer is option 2 

we will update the solution as soon as possible.

Concept:

Let F(x) be a cumulative distribution function then

(i) $\underset{x\to -\mathrm{\infty }}{lim}F(x)$ = 0, $\underset{x\to \mathrm{\infty }}{lim}F(x)$ = 1

(ii) F is non-decreasing function

Explanation:

(2): F(x) = $\{\begin{array}{ll}\frac{1}{2+{\mathrm{x}}^2}& \text{ if }\mathrm{x}<0,\\ \frac{2+{\mathrm{x}}^2}{3+2{\mathrm{x}}^2}& \text{ if }\mathrm{x}\ge 0\end{array}$

$\underset{x\to \mathrm{\infty }}{lim}F(x)$ = $\underset{x\to \mathrm{\infty }}{lim}\frac{2+x^2}{3+2x^2}$ = $\underset{x\to \mathrm{\infty }}{lim}\frac{2x}{4x}$ = $\frac{1}{2}$ (Using L'hospital rule), not satisfying

Option (2) is false

(3): F(x) = $\{\begin{array}{ll}\frac{1}{2+{\mathrm{x}}^2}& \text{ if }\mathrm{x}<0\text{, }\\ \frac{2\cos (\mathrm{x})+{\mathrm{x}}^2}{4+{\mathrm{x}}^2}& \text{ if }\mathrm{x}\ge 0\end{array}$

$\underset{x\to \mathrm{\infty }}{lim}F(x)$ = $\underset{x\to \mathrm{\infty }}{lim}\frac{2\cos x+x^2}{4+x^2}$ 

 = $\underset{x\to \mathrm{\infty }}{lim}\frac{-2\sin x+2x}{2x}$ (Using L'hospital rule)

= $\underset{x\to \mathrm{\infty }}{lim}(-\frac{\sin x}{x}+1)$ - 1 + 1 = 0, not satisfying

Option (3) is false

(4): F(x) = $\{\begin{array}{ll}\frac{1}{2+{\mathrm{x}}^2}& \text{ if }\mathrm{x}<0,\\ \frac{1+{\mathrm{x}}^2}{4+{\mathrm{x}}^2}& \text{ if }\mathrm{x}\ge 0\end{array}$

f(0^-) = 1/2 and f(0⁺) = 1/4 and $\frac{1}{2}>\frac{1}{4}$ so F(x) not satisfying the property "F is non-decreasing function"

Option (4) is false

Therefore option (1) is correct

Concept:

A parametric function f(λ) is said to be estimable if there exist g(X) such that E(g(X)) = f(λ) otherwise it is called not estimable.

Explanation:

Given X be a Poisson random variable with mean λ

So E(X) = λ and Var(X) = λ

We have to find the parametric function which is not estimable.

(2): E(X) = λ so here we are getting a function g(X) = X

So it is estimable

Option (2) is false

(3): E(X²) = [E(X)]² + Var(X) = λ² + λ

So E(X² - X) = E(X2) - E(X) =  λ2 + λ - λ = λ2 

Here we are getting the function g(X) = X2 - X

 So it is estimable

Option (3) is false

(4): E$(\sum \frac{(-1{)}^x{\lambda }^x}{x!})$ = e−λ​

Here we are getting the function g(X) = $(\sum \frac{(-1{)}^x{\lambda }^x}{x!})$

 So it is estimable 

Option (4) is false

Hence option (1) is correct

Concepts Used:

1. Cumulative Distribution Function (CDF):

The CDF F(x) gives the probability that the random variable X takes a value less than or equal to x . That is, F(x) = P(X ≤ x) .

2. Finding Probability Using the CDF:

The probability that the random variable X lies within a certain interval (a, b] is given by:
     
 P(a < X ≤ b) = F(b) - F(a)

3. Probability at a Specific Point (Jump Discontinuity):

The probability at a specific point x = c is the difference in the CDF just to the right and just to the left of c :
     
P(X = c) = F(c⁺) - F(c^-)

Explanation -

We are given a cumulative distribution function (CDF) F(x) of a random variable X as:

F(x) = $\{\begin{array}{ll}0,& \text{if }x<0\\ \frac{x+1}{3},& \text{if }0\le x<1\\ 1,& \text{if }x\ge 1\end{array}$

From the definition of the probability from the CDF:  P(a < X $\le$ b) = F(b) - F(a)

In this case, we need to calculate $F\left(\frac{3}{4}\right)andF\left(\frac{1}{3}\right)$.

Since $0\le \frac{1}{3}<1$  and $0\le \frac{3}{4}<1$ , we use the formula $F(x)=\frac{x+1}{3}$ for both 1/3  and 3/4 :

$F\left(\frac{1}{3}\right)=\frac{\frac{1}{3}+1}{3}=\frac{\frac{4}{3}}{3}=\frac{4}{9}$

$\Rightarrow F\left(\frac{3}{4}\right)=\frac{\frac{3}{4}+1}{3}=\frac{\frac{7}{4}}{3}=\frac{7}{12}$

Thus, the probability $P(\frac{1}{3}<X\le \frac{3}{4})$ is:

$P(\frac{1}{3}<X\le \frac{3}{4})=F\left(\frac{3}{4}\right)-F\left(\frac{1}{3}\right)=\frac{7}{12}-\frac{4}{9}$

= $\frac{21}{36}-\frac{16}{36}=\frac{5}{36}$

The probability at a point is the jump in the CDF at that point. We need to calculate F(0⁺) - F(0^-) .

From the CDF definition: F(0⁺) = F(0) =$\frac{0+1}{3}=\frac{1}{3}$

⇒ F(0^-) = 0

Thus, $P(X=0)=F(0^{+})-F(0^{-})=\frac{1}{3}-0=\frac{1}{3}=\frac{12}{36}$

Now, we add the two results:

$P(\frac{1}{3}<X\le \frac{3}{4})+P(X=0)=\frac{5}{36}+\frac{12}{36}=\frac{17}{36}$

Thus, the final answer is 17/36.

Explanation -

For a random variable X that follows a Uniform (0, 4) distribution, X will take any value between 0 and 4 with equal probability.

The defined $Y=\sqrt{(}X)$ will range from 0 to 2.

Option (1) - The range of Y is [0, 2]

 This statement is TRUE. $Y=\sqrt{(}X)$ will take a value between 0 (for X=0) and 2 (for X=4).

Option (2) - Both X and Y have the same mean  

This statement is NOT TRUE. In a uniform distribution, the mean is given by (a + b) / 2.

The mean of X will be (0+4)/2 = 2, and the mean of Y will be (0+2)/2 = 1, so they do not have the same mean.

Option (3) - The variance of X is greater than the variance of Y

This statement is TRUE. Variance measures the deviation from the mean, and since X can take on values from a wider range (0 to 4, as opposed to 0 to 2 for Y), it will generally have a greater variance.

Option (4) - The distribution of Y is also uniform

This statement is NOT TRUE. When we apply a transformation like $\sqrt{(}X)$, the resulting distribution Y will not retain the uniform property of X.

So, statements B and D are NOT true.

Hence the correct options are 2 and 4.