Hello! I am having trouble finding out how to solve letter D of this question, If anyone could help it would be greatly appreciated!

Question # 3

From a recent survey by a marketing company, it is found that the number of cars in Canadian households follows the following probability distribution.

# 0f CarsProportion of families

0.20

1.48

2.15

3.08

4.05

5.04.

a)Calculate the expected value and the variance of the number of cars in Canadian households.

E(X) = ( (0x0.2) + 1(0.48) + 2(0.15) + 3(0.08) + 4(0.05) + 5(0.04) )

E(X) =( 0.48 + 0.3 + 0.24 + 0.20 + 0.20 )

[ E(X) = 1.42 ]

= P(X)(X - )

= { 0.2(0-1.42)²+ 0.48(1-1.42)²+ 0.15(2-1.42)²+ 0.08(3-1.42)²+ 0.05(4-1.42)²+ 0.04(5-1.42)²}

= { 0.2(2.0164) + 0.48(0.1764) +0.15(0.3364) + 0.08(2.4964) + 0.05(6.6564) + 0.04(12.8164)

[ = 1.5836]

b)What is the probability of a randomly selected household having 2 or 3 cars?

P(2 or 3)= Sum of probabilities of 2 & 3 car.

So P(2 or 3) = P(2) + P(3)

[ P(2 or 3) = 0.23 ]

c)Calculate the probability of a randomly selected household havingat least 2 cars.

P(At least 2) =P(At least 2) = 1-(P(0) + P(1) )

P(X 2) = 1- (0.2 +0.48)

P(X 2) = 0.32

d)If 5 households are randomly selected, what is the probability that 3 of them having 2 or 3 cars?

house 1

X~ bin( n=5, p= 0.23)

e)What is the probability that at least one household having 2 or 3 cars?

P(At least one house w. 2 or 3 cars) = 1 - P(1-0.23)⁵

[P(At least one house w. 2 or 3 cars) = 0.7293]