Question

It has been estimated that only about 15% of Illinois residents regularly recycle. Suppose you randomly survey 100 residents. We are interested in the number who recycle. Identify the following variables:

n =

P =

q =

x =

I. What is the distribution. Support your answer by verifying all applicable criteria.

II. Write the first 3 rows of the PDF

III. What is the probability that at least 8 residents recycle

IV. Is this a symmetric distribution? Why or why not?

V. Find the mean and standard deviation.

VI. In groups of 100 randomly selected residents, what is the expected value? Explain what that means.

Answer 1

n=100=Number of observations/trials.

P=15/100=0.15=probability of success.

q=(1-0.15)=0.85=Probability of failure.

x=Number of residents who recycle, x=0,1,2,......,100.

I. The distribution is Normal Distribution .

The conditions are:

The number of trials=n=100 is fixed.

Each trial is independent.

Each trial has one of these 2 outcomes:Residents do recycle (Success )and Residents do not recycle (Failure).

For each trial ,The probability of success=0.15 is same.

Hence, X follows Binomial(100,0.15).

II. This is a pmf not a pdf.

So,

Mass points are 0,1,2,...100.

The first 3 rows of the PMF are:

P(X=x)=??? *0.15^x*(1-0.15)^(100-x) ,x=0,1,2,.....,100.

f(x)=P(X=x)=?? *0.15^x*0.85^(100-x),x=0,1,2,...,100.

=[100!/{(x)! *(100-x)!}]*0.15^x*0.85^(100-x) ,x=0,1,2,...,100

=[{100*(100-1)*...*(100-x-1)}/x!)]*0.15^x*0.85^(100-x) ,x=0,1,2,...,100.

III.

P(X?8)=1-P(X?7)=1-[P(X=0)+P(X=1)+....+P(X=7)]=1-0.01216519191.=0.98783480809.

IV.No,here Binominal Distribution is not symmetric because p=0.15

For Binominal distribution to be symmetric, p must be 1/2 or 0.5.

V. Mean=100*0.15=15.

Standard deviation=sqrt(100*0.15*0.85)=3.57071421.

VI. Expected value is the mean of the random variable.

E(X)=?? ???x*P(X=x)=0*P(X=0)+1*P(X=1)+

....+100*P(X=100).

E(X)=100*0.15=15.