Mean and Variance of Random variables MCQ Quiz - Objective Question with Answer for Mean and Variance of Random variables - Download Free PDF

Concept:

Probability Distribution:

• A probability distribution describes how the probabilities are distributed over the values of the random variable.
• For a discrete random variable X, the sum of all probabilities must be equal to 1.
• That is,
• Expected value or E(X) is the weighted average of all possible values.
• Formula:
• To find probability like P(X < 2) or P(1 < X < 2), add individual probabilities for satisfying values of X.

Calculation:

Given,

P(X=0) = k, P(X=1) = 2k, P(X=2) = 3k

⇒ Total Probability = k + 2k + 3k = 6k

⇒ 6k = 1

⇒ k = 1 / 6

⇒ P(X < 2) = P(X = 0) + P(X = 1)

⇒ P(X < 2) = k + 2k = 3k = 3 × (1/6) = 1/2

⇒ E(X) = 0 × k + 1 × 2k + 2 × 3k

⇒ E(X) = 0 + 2k + 6k = 8k = 8 × (1/6) = 4/3

⇒ P(1 < X < 2) = P(X = 2) = 3k = 3 × (1/6) = 1/2

∴ Correct statements are (A) and (B) only.

Calculation

Probability distribution

\(\begin{array}{c|c} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \\ \hline \mathrm{x}=0 & \frac{7 \mathrm{C}_{2}}{10 \mathrm{C}_{2}}=\frac{42}{90} \\ \hline \mathrm{x}=1 & \frac{7 \mathrm{C}_{1} \times 3 \mathrm{C}_{1}}{10 \mathrm{C}_{2}}=\frac{42}{90} \\ \hline \mathrm{x}=2 & \frac{3 \mathrm{C}_{2}}{10 \mathrm{C}_{2}}=\frac{6}{90} \end{array}\)

Now, 

\(\mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{42}{90}+\frac{12}{90}=\frac{54}{90}\)

\(\sigma^{2}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{1}^{2}-\mu^{2}=\frac{42}{90}+\frac{24}{90}-\left(\frac{54}{90}\right)^{2}\)

⇒ \(\frac{66}{90}-\left(\frac{54}{90}\right)^{2}\)

⇒ \(\sigma^{2} \Rightarrow \frac{28}{75} \)

Hence option 1 is correct

Calculation 

HHH → 0

HHT → 0

HTH → 1

HTT → 0

THH → 1

THT → 1

TTH → 1

TTT → 0

Probability distribution

\(\begin{array}{c|c|c} \mathrm{x}_{\mathrm{i}} & 0 & 1 \\ \hline \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right) & 1 / 2 & 1 / 2 \end{array}\)

\(\mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{2}\)

\(\sigma^{2}=\sum x_{i}^{2} p_{i}-\mu^{2}\)

= \(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

\(64\left(\mu+\sigma^{2}\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48\)

Hence option 2 is correct

Calculation

E(X) =  \(P(X)\cdot \frac{n}{2}\cdot (n+1) =\frac{1}{6}\cdot \frac{6}{2}\cdot 7 =\frac{7}{2}\)

E(X²) = \(\frac{1}{6}\times \frac{6}{6} \times 7\times 13\) =  \(\frac{91}{6}\)

Var.(X) = {E(X)}² - E(X2) = \(\frac{49}{4}-\frac{91}{6} = \frac{35}{12}\)

Hence option 3 is correct

Concept:

Mean (Expected Value) of a Discrete Random Variable:

• The mean or expected value of a discrete random variable X is the weighted average of all possible values that X can take.
• It is calculated as: E(X) = Σ [x × P(X = x)]
• All probabilities must sum up to 1: Σ P(X = x) = 1
• The result of E(X) represents the average outcome over many repetitions of the random experiment.

Calculation:

Given,

Probability distribution is:

Step 1: Find the total probability

Sum of all probabilities = 1

K/10 + 2/10 + K/10 + 3/10 + (K + 2)/10 + K/10 = 1

⇒ (4K + 7)/10 = 1

⇒ 4K + 7 = 10

⇒ 4K = 3

⇒ K = 3/4

Step 2: Compute the expected value E(X)

E(X) = Σ [x × P(X = x)]

E(X) = (-2 × 3/40) + (-1 × 2/10) + (0 × 3/40) + (1 × 3/10) + (2 × 11/40) + (3 × 3/40)

⇒ E(X) = -6/40 - 8/40 + 0 + 12/40 + 22/40 + 9/40

⇒ E(X) = (-6 - 8 + 12 + 22 + 9)/40

⇒ E(X) = 29/40

⇒ E(X) = 15/15 - 14/15 = 4/5

∴ Mean of X is 4/5.

Calculation:

Probability Distribution of Z is

0.4 + k + 2k = 1

3k = 1 - 0.4

k = 0.2

P(you study atleast two hours) 

P(Z ≥ 2) = P(2) + P(3) + P(4) + .....

              = 2k

              = 2 × 0.2

P(Z ≥ 2) = 0.4

Concept:

For a random variable X = xi with probabilities P(X = x) = pi:

• Mean/Expected Value: μ = ∑pixi.
• Variance: Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2.
• Standard Deviation: σ = \(\rm \sqrt{Var(X)}\).

Calculation:

We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3.

Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3)

= 0.3 + 0.8 + 0.9

= 2

And, ∑pi(xi)2 = (0.3 × 1²) + (0.4 × 2²) +(0.3 × 3²)

= 0.3 + 1.6 + 2.7

= 4.6

∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6.

The variance of the distribution is Var(X) = 0.6.

Formula used:

Variance = npq

Where,

p = probability of success,

q = the probability of failure,

n = the total no. of attempts

Calculation:

p (probability of drawing a blue balloon) = \(\rm \frac{10}{15} = \frac{2}{3} \)

Here, n = 5, p = \(\rm \frac{2}{3} \) ,q = \(\rm \frac{1}{3} \)

Then, variance = \(\rm 5 \times \frac{2}{3} \times \frac{1}{3} \) = \(\rm \frac{10}{9} \)

The variance of the number of blue balloons drawn is \(\rm \frac{10}{9} \)

Concept:

Expected Value (or mean) of a Discrete Random Variable:
For a discrete random variable, the expected value usually denoted as μ or E(X), is calculated using

μ = E(X) = ∑ x_iP(x_i)

1. The formula means that we multiply each value x, in the support by its respective probability P(x) and then add them all together.

2. It can be seen as an average value but weighted by the likelihood of the value.

Calculation:

We know that, mean

μ = E(X) = ∑ x_iP(x_i)

⇒ E(X) = 1 × 0.2 + 2 × 0.35 + 3 × 0.25 + 4 × 0.15 + 5 × 0.05

⇒ E(X) = 0.2 + 0.7 + 0.75 + 0.6 + 0.25

⇒ E(X) = 2.5

Additional Information
The variance of a discrete random variable is given by:

Var(X) = σ²(X) = E(X²) - [E(X)]²

The formula means that first, we sum the square of each value times its probability then subtract the square of the mean.

Concept:

The expectation of a random variable E(X²) is given by \(\Sigma {x^2P(x) }\)

Calculation:

Given 

Expectation E(X²) is calculated by \(\Sigma {x^2P(x) }\)

⇒ E(X²) = \(\left(1^2\times\frac{1}{10}\right) +\left(2^2\times\frac{1}{5}\right)+\left(3^2\times\frac{3}{10}\right)+\left(4^2\times\frac{2}{5}\right)\)

⇒ E(X²) = \(\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}\)

⇒ E(X²) = \(\frac{100}{10}\) = 10 

The value of E(X²) is equal to 10.

The correct answer is option 4.

Calculation:

Let X be the random variable representing a number on the die.

The total number of observation is ten.

P(X = 5) = \(\rm \frac{4}{10} = \frac{2}{5}\) 

P(X = 4) = \(\rm \frac{3}{10} \)

P(X = 3) = \(\rm \frac{2}{10} = \frac{1}{5}\)

P(X = 2) = \(\rm \frac{1}{10} \)

Mean(X) = \(\rm \sum x_{i}p_{i}\)

Mean(X) = \(\rm 5 \times \frac{2}{5} + 4 \times \frac{3}{10} + 3 \times \frac{1}{5} + 2 \times \frac{1}{10} \)

Mean(X) = \(2 + \rm \frac{20}{10} =4 \)

Concept

When a fair dice is rolled, probability of getting any number is \(\frac{1}{6}\).

Suppose an experiment is repeated n times. Probability of success be p.

If X denote the number of success, then

P(X = k) = C(n, k) × p^k (1 – p)^{n - k} , 0 ≤ k ≤ n

where

\({\bf{C}}\left( {{\bf{n}},\;{\bf{k}}} \right) = \;\frac{{{\bf{n}}!}}{{{\bf{k}}!\left( {{\bf{n}} - {\bf{k}}} \right)!}}\)

Mean of X = np

Standard deviation of X \(= \sqrt {{\bf{np}}\left( {1 - {\bf{p}}} \right)}\)

Calculation

A dice is rolled 8 times so, n = 8.

In a single roll of dice, probability of getting 5 or 6 is

= probability of getting 5 + probability of getting 6

\(= \frac{1}{6} + \frac{1}{6}\)

\(= \frac{1}{3}\)

So probability of success \({\bf{p}} = \frac{1}{3}\)

\({\bf{Mean}} = {\bf{n}} \times {\bf{p}} = \frac{8}{3}\)

\({\bf{Standard}}\;{\bf{deviation}} = \sqrt {{\bf{np}}\left( {1 - {\bf{p}}} \right)} = \frac{4}{3}\)

Correct option is (2).

Concept:

Mean = E[X] 

\(E\left[ X \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } x.{f_x}\left( x \right)dx\)

\(E\left[ {{X^2}} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {x^2}{f_x}\left( x \right)dx\)

Where f_x(x) is the probability density function.

Analysis:

\(Var\left( X \right) = E\left[ {{{\left( {X - {\mu _x}} \right)}^2}} \right]\)

\( = E\left[ {{X^2} - 2{\mu _x}X + \mu _x^2} \right]\)

\( = E\left[ {{X^2}} \right] - 2E\left[ {{\mu _x}X} \right] + E\left[ {\mu _x^2} \right]\)

\( = E\left[ {{X^2}} \right] - 2{\mu _x}E\left[ X \right] + \mu _x^2\)

\( = E\left[ {{X^2}} \right] - 2\mu _x^2 + \mu _x^2\)

\( = E\left[ {{X^2}} \right] - \mu _x^2\)

Since, \( E[X]= \mu _x\)

\(Var\left[ X \right] = E\left[ {{X^2}} \right] - {E^2}\left[ X \right]\)

Concept:

The expectation of a random variable E(X) is given by \(\Sigma {xP(x) }\)

Calculation:

Given 

Expectation E(X) is calculated by \(\Sigma {xP(x) }\)

⇒ E(X) = (- 4) × 0.1 + (- 3) × 0.2 + (- 2) × 0.3 + (- 1) × 0.2 + 0

⇒ E(X) = - 0.4 - 0.6 - 0.6 - 0.2

⇒ E(X) = - 1.8

The value of E(X) is  -1.8.

The correct answer is option 4.

Concept:

In B (n, p). n is number of observations.

mean = np

variance = npq 

Calculations:

Given, a r.v.X ∼ B (n, p). n is a number of observations.

If values of mean and variance of X are 18 and 12 respectively

mean = np = 18 ....(1)

variance = npq = 12

⇒\(\rm \dfrac{npq}{np}= \dfrac{12}{18}\)

⇒ q = \(\rm \dfrac 2 3\) 

we know that p = 1 - q

⇒ p = 1 - \(\rm \dfrac 2 3\)

⇒ p = \(\rm \dfrac 1 3\)

Equation (1), we have

⇒n\(\rm \times\dfrac 1 3\) = 18

⇒n = 54

⇒n = 0, 1, 2, ....54

Hence, the number of observations = 55