Question

In an annual poll about consumption habits, telephone interviews were conducted for a random sample of 1,016 adults aged 18 and over. One of the questions was, "How many cups of coffee, if any, do you drink on an average day?" The following table shows the results obtained.

Number of Cups per Day	Number of Responses
0	365
1	264
2	193
3	91
4 or more	103

Define a random variable x = number of cups of coffee consumed on an average day. Let

x = 4

represent four or more cups.

(a)

Develop a probability distribution for x. (Round your answers to three decimal places.)

Number of Cups per Day	f(x)
0
1
2
3
4 or more

(b)

Compute the expected value of x. (Round your answer to three decimal places.)

E(x) =

(c)

Compute the variance of x. (Round your answer to three decimal places.)

Var(x) =

(d)

Suppose we are only interested in adults who drink at least one cup of coffee on an average day. For this group, let y = the number of cups of coffee consumed on an average day. Compute the expected value of y. (Round your answer to three decimal places.)

E(y) =

Compare the expected value of y to the expected value of x.

When we only take into account adults that drink at least one cup of coffee per day, the expected value is  ---Select--- equal to higher than lower than the expected value of x.

Answer 1

a)

dividint each value with total frequency:

x	P(X=x)
0	0.359
1	0.260
2	0.190
3	0.090
4	0.101

b)

x	P(X=x)	xP(x)	x2P(x)
0	0.359	0.00000	0.00000
1	0.260	0.25984	0.25984
2	0.190	0.37992	0.75984
3	0.090	0.26870	0.80610
4	0.101	0.40551	1.62205
total		1.3140	3.4478
	E(x) =μ=	ΣxP(x) =	1.3140
	E(x2) =	Σx2P(x) =	3.4478
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.721301
	std deviation=	σ= √σ2 =	1.31198

E(X) =1.314

c)

Var(x) =1.719 (please try 1.721 if this comes wrong and reply)

d)

x	P(X=x)	xP(x)
1	0.406	0.40553
2	0.296	0.59293
3	0.140	0.41935
4	0.158	0.63287
total		2.0507

from above E(Y) =2.050 (please try 2.051 if this comes wrong)

When we only take into account adults that drink at least one cup of coffee per day, the expected value is   higher than the expected value of x.