Question

2. According to a pet census, in a large city, the probability that a family will have no more than x pets is given by 2 4 Pino more .29 than x pets) .63 94 .98 P(exactly pets) No one in the census had more than 5 pets a) (5 pts.) Fill in the boxes for the probabilities of exactly x pets per family. b) (5 pts.) Calculate the average number of pets per family ECx.

Answer 1

a) Here we are given p ( no more than x ) i.e. p ( X $\leq$ x ) = F (x)

So to find p ( x) we subtract previous probability from the corresponding probability.

So for x = 0

p ( x ) = F (x) = 0.29

For x = 1 ,

p ( x ) = 0.63 - 0.29 = 0.34

For x = 2

p ( x) = 0.88 -0.63 = 0.25

For x = 3,

p ( x) = 0.94 -0.88 = 0.06

For x =4

p (x ) = 0.98 - 0.94 = 0.04

x	0	1	2	3	4
p(no more than x pets )	0.29	0.63	0.88	0.94	0.98
p ( exactly x pets )	0.29	0.34	0.25	0.06	0.04

b)

E(x) = $\sum x*p(x)$

= 0 * 0.29 + 1 * 0.34 + 2* 0.25 + 3* 0.06 + 4 * 0.04

= 1.18