According to the California Water Agency, 26% of Californians actively conserve

ID: 3041746 • Letter: A

Question

According to the California Water Agency, 26% of Californians actively conserve water. Three Californians are chosen at random.

a) Find the probability that all three actively conserve water.

b) Find the probability that exactly two actively conserve water by summing the probability of each

successful outcome.

c) Fifty Californians are now chosen at random. Do you think the method you used to calculate 4B

should be used to calculate the probability that exactly twelve Californians actively conserve water?

Research has shown that 20% of computers have malware. Suppose 11 computers are randomly selected.

a) Define the random variable of interest, X.

b) State the distribution of X. Show your work by checking the relevant criteria.

c) What is the probability that none of them have malware?

d) What is the probability that at least 2 computers have malware?

e) What is the probability that between (inclusive) 3 and 5 computers have malware?

f) Determine the mean.

Ans:

Binomial distribution:

n=3,p=0.26

P(x=k)=3Ck*0.26k*0.743-k

a)P(x=3)=0.263=0.0176

b)P(x=2)=3C2*0.262*0.741=0.1501

c)Now,n changes to 50,so probabilty distn. function will change.

P(x=12)=50C12*0.2612*0.7438

We can also do normal approximation,as formula above will be too lengthy to solve by hand,as n is large here.

a)x=number of computers having malware

b)Binomial distribution(n=11,p=0.2),as independent trials and only two possible outcomes.

c)P(x=0)=11C0*0.20*0.811=0.0859

d)P(x>=2)=1-P(x<2)=1-P(x=0)-P(x=1)=1-0.0859-11C1*0.21*0.810

=1-0.0859-0.2362

=0.6779

e)P(3<=x<=5)=P(x=3)+P(x=4)+P(x=5)=0.2215+0.1107+0.0388=0.3709

f)mean=np=11*0.2=2.2

x p(x) 0 0.0859 1 0.2362 2 0.2953 3 0.2215 4 0.1107 5 0.0388 6 0.0097 7 0.0017 8 0.0002 9 0.0000 10 0.0000 11 0.0000

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