Statistics & Exploratory Data Analysis MCQ Quiz - Objective Question with Answer for Statistics & Exploratory Data Analysis - Download Free PDF

The Correct Answer is option 3 

We are Mainly focused on the Pure and Applied Mathematics 

We will try to provide the solution of Statistics Part if Possible 

Concept:

The maximum likelihood estimators of the mean of normal distribution is 

Explanation:

Given a random sample {3, 6, 9} of size 3 from a normal distribution with mean μ and variance 1. 

So maximum likelihood estimate of μ =  =  = 6

But given μ ∈ (−∞, 5] so maximum likelihood estimate of μ can't be 6.

Now since 5 is the maximum value in (−∞, 5]

So maximum likelihood estimate of μ is 5

Option (2) is correct   

Concept: 

Maximum Likelihood Estimation (MLE), which is a method for estimating the parameters of a statistical model given observed data.

The likelihood function is the product of the probabilities of each observed outcome.

Explanation:

where  is an unknown parameter.

The sample size is 100.

The observed counts of x = 0, 1, 2 are 20, 30, and 50, respectively.

Let's denote the observed counts as

 for 

 for 

 for 

The likelihood function is the product of the probabilities of the observed data points, based on the PMF.

The probability of observing x = 0, 1, 2 given   is

The constant terms involving  can be ignored when maximizing the function.

To find the MLE, take the derivative of  with respect to  and set it equal to zero

For maximum value: 

⇒ 

Solving the equation: 

⇒ 

⇒ 

⇒ 

Substitute  and 

So, the correct option  is 2).

Concept:

Let F(x) be a cumulative distribution function then

(i)  = 0, = 1

(ii) F is non-decreasing function

Explanation:

(2): F(x) = 

 =  =  =  (Using L'hospital rule), not satisfying

Option (2) is false

(3): F(x) = 

 =  

 =  (Using L'hospital rule)

=  - 1 + 1 = 0, not satisfying

Option (3) is false

(4): F(x) = 

f(0^-) = 1/2 and f(0⁺) = 1/4 and \frac14\) 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 = e−λ​

Here we are getting the function g(X) = 

 So it is estimable 

Option (4) is false

Hence option (1) is correct

Concept:

Covariance Matrix Calculation:
 

Each  and since the 's are i.i.d., their variances and covariances can be computed easily.
 

The variance of  is 
  
The covariance between any two distinct  and  (where  ) is

Thus, the covariance matrix of  Y has 2's on the diagonal and 1's off-diagonal.

Explanation:

  are independent and identically distributed (i.i.d.) random variables with mean 0 and variance 1.

 .

The task is to find the first principal component based on the covariance matrix of  .

Each  where,  is common across all 's.

The covariance between any two different  and  depends on  .

The covariance matrix  for  will have entries:

, where  is the Kronecker delta.

Since  and  have variance 1, we get,

 when  (because of the common ).

  when .

Thus, the covariance matrix  is a matrix with diagonal entries 2 and off-diagonal

entries 1. It is a symmetric matrix.

The first principal component corresponds to the eigenvector associated with the

largest eigenvalue of the covariance matrix  .

For a covariance matrix like this (with all off-diagonal elements equal and diagonal elements

greater than off-diagonal elements), the first principal component will have equal weights on all

components. Specifically, the eigenvector corresponding to the largest eigenvalue will be

proportional to .

The first principal component can thus be expressed as ​​

This is a linear combination of the 's, where each  has an equal weight,

scaled by  to ensure unit length of the eigenvector.

 
From the available options, the correct representation of the first principal

component is ​​

Thus, the correct answer is the first option.

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

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) =

From the definition of the probability from the CDF:  P(a  b) = F(b) - F(a)

In this case, we need to calculate .

Since   and , we use the formula for both 1/3  and 3/4 :

Thus, the probability is:

= 

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) =

⇒ F(0^-) = 0

Thus, 

Now, we add the two results:

Thus, the final answer is 17/36.

Concept:

Mean vector of Y =  is E(Y) =  and covariance matrix of Y is Σ = 

Explanation:

Given mean vector of  X = (X1, X2)T is (0, 0)T and covariance matrix Σ such that Σ = 

So 

E(X1) = 0, E(X2) = 0, var(X1) = 5, var(X2) = 10........(i)

cov(X1X2) = cov(X2X1) = -3................................(ii)

Now, cov(X1X2) = -3

⇒ E(X1X2) - E(X1)E(X2) = -3

⇒ E(X1X2) - 0 = -3 (using (i))

⇒ E(X1X2) = -3 ..........(iii)

Y = (X1, 5 − 2X2)T

Let Y = (Y₁, Y₂)^T

Then Y1 = X1 and Y2 = 5 − 2X2.......(iv)

Now, E(Y1) = E(X1) = 0 and E(Y2) = 5 - 2E(X2) = 5 - 0 =5 

So mean vector of Y is  = .......(v)

Also var(Y1) = var(X1) = 5

var(Y2) = var(5 - 2X2) = 0 + 4var(X2) = 4 × 10 = 40

cov(Y1Y2) = E(Y1Y2) - E(Y1)E(Y2)

= E(5X1 - 2X1X2) - 0 (using (i) and (ii))  

= 5E(X1) - 2E(X1X2) = 0 + 6 = 6 (using (i) and (iii))

Also cov(Y2Y1) = 6

Therefore covariance matrix of Y is 

Therefore option (4) is correct.

Explanation:

Given X1, X2, …, Xn are independently and identically distributed N(θ, 1) 

X̅ = n−1  Xi 

t0.975, n−1 denote the 0.975-quantile of a Student's− t distribution with n − 1 degrees of freedom.

So here confidence interval is 0.975 which is  more than 0.95.

Therefore level of the significance = 2.25%

Hence option (3) is correct