Question

According to the CDC, approximately 30% of gonorrhea cases tested in 2011 were resistent to at least three major antibiotics commonly used to treat gonorrhea. A physician treated 12 cases of gonorrhea during one week in 2011.

a) What is the approprate distribution to model the data, and why?

b) What is the probablity that exactly three of the 12 cases were resistent to at least one of three major antibiotics?

c) What is the probablity that at most three of the 12 cases were resistent to at least one of three major antibiotics?

d) What is the probibilty that at least ten of 12 cases were resistent to at least one of the three major antibiotics?

e) Compute the mean and standard deviation of number of successes in this problem.

Answer 1

Sol:

n=12

p=prob of success=0.30

q=prob of failre=1-p=1-.30=0.70

np=12*0.3=3.6

nq=12*0.7=8.4

X follows normal distributio with n=12 and p=0.30

a) What is the approprate distribution to model the data, and why?

successes =np=3.6

failures=nq=8.4

are less than 10%

fixed number of trials n=12

the probability of success remains constant from trial to trial.

b) What is the probablity that exactly three of the 12 cases were resistent to at least one of three major antibiotics?

P(X=3)

n=12

p=0.3

q=0.7

P(X=x )=ncx*p^x*q^n-x

=12C3*0.30^3*0.70^12-3

=0.2397

Solutionc:

P(X<3)

=0.2528

Solutiond:

P(X>=10)

=P(X=10)+P(X=11)+P(X=12)

=0.00019+0.000015+0.000001

=0.0002

ANSWER:0.0002

e) Compute the mean and standard deviation of number of successes in this problem.

Mean=np=12*0.3=3.6

standard deviation =sqrt(npq)

=sqrt(12*0.3*0.7)

=1.5875

ANSWER:MEAN=3.6

standard deviation=1.5875