Solution

Arrange the data in Ascending order

2, 3, 4, 7 , 9 and 11

Median of even number of terms = (n + n + 1)/2 = (4 + 7)/2 = 11/2 = 5.5

Solution

Coefficient of variance = (SD/M) × 100

(SD/M) × 100

⇒ (10/22) × 100

⇒ (5/11) × 100

⇒ 500/11

⇒ 45.45

⇒ Coefficient of variance = 45.45%

Solution

Number of observations = 6

Mean of X = 17.67

Mean of Y = 25.5

Cov(X, Y) = (⅙) [(2 – 17.67)(5 – 25.5) + (8 – 17.67)(12 – 25.5) + (18 –

17.67)(18 – 25.5) + (20 –

17.67)(23 – 25.5) + (28 – 17.67)(45 – 25.5) + (30 – 17.67)(50 – 25.5)]

= 157.83

Solution

The correlation coefficient between two variables X and Y = 0.4

The correlation coefficient (r) is independent of origin and scale and depend on

the sign of variables

The correlation coefficient between the two variables is the measure of the

slope between the

variables in the regression graph. It is given that the correlation coefficient

between X and Y is 0.4 and the correlation coefficient is independent of change

of origin and scale but it depends on variables

The correlation coefficient between 2X and (-Y) is - 0.4

Solution

Solution

Given:

Marks of students: 85, 65, 55, 42, 69, 83, 92, 77.

We know that,

Thus the standard deviation for this data is 15.63

Solution

The ascending order of given observations is,

127, 147, 153, 153, 153, 161, 165, 165, 168, 170, 179, 180, 185, 187, 189, 190,

190, 194

Here, n = 18

Median = 1/2 [(n/2) + (n/2 + 1)]th observation

= 1/2 [9 + 10]th observation

= 1/2 (168 + 170)

= 338/2

= 169

Thus, the median is 169

Solution

A. Var[X+Y] = Var[X] + Var[Y] + 2∙Cov[X,Y] Above formulae clearly shows that Var(X+Y) depends on covariance which in turn depends on independence.

B. Always holds.

C. It will be Var(X-Y) = Var(𝑋)+Var(𝑌)−2Cov(𝑋,𝑌) because Var(−𝑌)=Var(𝑌). For independent case, Var(X-Y)= Var(X) + Var(Y)

D. It’s a standard axiom.

Solution

A.Not necessarily true.

B. If E[X]>E[Y], it implies that on an average, X takes on larger values than Y. Therefore, there must be some values that X can take which are larger than those that Y can take. Hence, there’s a non-zero chance (positive probability) that a realization of X will be greater than a realization of Y in some outcomes within their shared probability space. Hence P[x>y]>0 is correct.

C. These two probabilities, p(A | B) and p(A | Bc), are not necessarily equal. They can be equal in some specific cases, but it’s not a general rule. The probabilities depend on the specific relationship between events A and B in the sample space.

D.This is Jensen’s inequality which always holds since it comes from formulae of variance and variance is always non-negative. Hence, Correct.

Solution

Solution

Solution

Use P (max(X, Y ) > 3) = 1 − P (max(X, Y ) ≤ 3) = 1 − P (X ≤ 3) P (Y ≤ 3),

and plug in the values for P (X ≤ 3),

P (Y ≤ 3) use their distributional information to arrive at the answer.

Solution

Solution

Given the observations 2, 3, 5, 6, x, 8, 9, to determine the median, we arrange the observations in ascending order: 2, 3, 5, 6, x, 8, 9.

Since the number of observations is odd, the median is the middle value in the ordered list. In this case, the fourth observation, which is 6, represents the median.

To identify the values that x cannot be equal to, we need to consider the possible positions of x in the ordered list.

If x is less than 6, it would be positioned to the left of 6 in the list. In such a case, the median would be greater than 6, as the middle value would shift to a higher number. Therefore, x cannot be less than 6

If x is greater than 6, it would be positioned to the right of 6 in the ordered list. In this case, the median would still be 6 since it represents the middle value. Therefore, x can be greater than 6.

If x equals 6, it would coincide with the median, as it represents the middle value in the ordered list. Therefore, x can be equal to 6.

Hence, considering the given information that the median is known to be 6, it can be concluded that x cannot be equal to 1, as it is less than 6.

Solution

Given:

Calculation: