Test: Engineering Mathematics- 9

Mode refers the most frequently appeared value (which is 17 in this case).

If X is random variable, then Poisson’s distribution is given by:

X ~ U(6,12)

Comparing with X ~ U(a,b)

∴ a = 6 and b = 12
Mean of uniform distributed variable is given as

G(-1) = p(-2) + p(-1)

         = 0.2 + 0.2 = 0.4

G(1) = p(-2) + p(-1) + p(0) + p(1)

        = 0.2 + 0.2 + 0.2 + 0.2

        = 0.8

P(success) = Probability of getting 2 or 5 is 2/6 = 1/3

P(failure) = Probability of not getting 2 or 5 is 4/6 = 2/3
Probability of 0 success and 3 failure is 
Probability of 1 success and 2 failure is 

Probability of 2 success and 1 failure is   
Probability of 3 success and 0 failure is

x is the number of success and p(x) is probability of x times success.

q² + 2pq + p² = 1

(p+q)² = 1

∴ (p+q) = 1

= 0 × q² + 1 × 2pq + 2 × p²

= 2p (q+p)

∴ μ = 2p

 

Var(x) = 2pq + 4p² − 4p²

Var(x) = 2pq

Mean = 4

k ≠ m so k = m = 4 is out.

{k, m} = {1, 7} or {2, 6} or {3, 5}

for median of {3, k, 2, 8, m, 3}

{1, 2, 3, 3, 7, 8} or {2, 2, 3, 3, 6, 8} or {2, 3, 3, 3, 5, 8}

Median: (3 + 3) /2 = 3

Var(X) = E(X) = 6

Var(X) = E(X²) - (E(X))²

6 = E(X²) – 36

∴ E(X²) = 42

[E(X + 3)²]

= E(X² + 6X + 9 )

= E(X²) + 6E(X) + 9         

= 42 + 6× 6 + 9

= 87  

Since y is a random variable:

= 0 + m+ m + 3m + 3m + m² + 4m² + 5m² + m = 1

10m² + 9m−1 = 0

(m+1) (10m−1) = 0

∴ m = −1or m = 1/10
since probability cannot be in negative

np = 4 and npq = 3

thus q = (3/4) and p = (1-q) = (1/4)

mode is an integer such that np+p > x > np-q

⇒ 4 +(1/4) > x >4-(3/4)

⇒ (13/4)<x<(17/4)

3.25<x<4.25

⇒ x = 4