Question

2. In a large scavenger hunt, the number of people on each team is distributed as follows:

   A team is selected.

   1. a) What is the probability that there is only one person on the team?

   2. b) What is the probability there are between 3 and 6 people on the team?

   3. c) What is the average number of people on a team?

   4. d) Determine the standard deviation.

x. 2. 4. 5. 6. 8

P(X=x). 0.15. 0.44. 0.26. 0.10 0.05

Answer 1

a) probability that there is only one person on the team =P(X=1)=0

(as from given distribution x=1 does not exist)

b) probability there are between 3 and 6 people on the team =P(3<=X<=6)=P(X=5)+P(X=6)=0.26+0.1

=0.36

c)

x	P(x)	xP(x)	x2P(x)
2	0.15	0.300	0.600
4	0.44	1.760	7.040
5	0.26	1.300	6.500
6	0.1	0.600	3.600
8	0.05	0.400	3.200
	total	4.360	20.940
	E(x) =μ=	ΣxP(x) =	4.3600
	E(x2) =	Σx2P(x) =	20.9400
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.930
	std deviation=	σ= √σ2 =	1.389

average number of people on a team =4.36

d)

standard deviation =1.389