Question

Standard deviation for fruits and vegetables

The following table gives the distribution of the number of servings of fruits and vegetables consumed per day in a population.

Number of servings X	0	1	2	3	4	5
Probability	0.3	0.1	0.1	0.2	0.2	0.1

Find the standard deviation for the distribution of the number of servings of fruits and vegetables. (report to 1 decimal place)

Answer 1

x	P(x)	xP(x)	x2P(x)
0	0.3	0.000	0.000
1	0.1	0.100	0.100
2	0.1	0.200	0.400
3	0.2	0.600	1.800
4	0.2	0.800	3.200
5	0.1	0.500	2.500
	total	2.200	8.000
	E(x) =μ=	ΣxP(x) =	2.2000
	E(x2) =	Σx2P(x) =	8.0000
	Var(x)=σ2 =	E(x2)-(E(x))2=	3.1600
	std deviation=	σ= √σ2 =	1.778

standard deviation for the distribution of the number of servings of fruits and vegetables =1.8