Question
Medical treatment will cure about 86% of all people who suffer from a certain eye disorder. Suppose a large medical clinic treats 57 people with this disorder. Let r be a random variable that represents the number of people that will recover. The clinic wants a probability distribution for r. Write a brief but complete description in which you explain why the normal approximation to the binomial would apply. Are the assumptions satisfied? Explain. Estimate P(r lessthanorequalto 47). Estimate P(47 lessthanorequalto r lessthanorequalto 55). The diameters of oranges from a Florida orchard are normally distributed with mean mu = 3.2 inches and standard deviation sigma = 1.1 inches. A packing supplier is designing special occasion presentation boxes of oranges and needs to know the average diameter for a random sample of 8 oranges. What is the probability that the mean diameter F for a sample of 8 oranges is smaller than 3 inches? larger than 3.5 inches? between 3.1 and 3.3 inches? The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is mound shaped and symmetrical with mean mu = 25.9 mpg and standard deviation sigma = 9.5 mpg. If 30 such cars are tested, what is the probability the average mpg x is less than 23 mpg? greater than 28 mpg? between 23 and 28 mpg?
Explanation / Answer
9.Using Central limit theorem the sampling distribution of sample mean is also normal with mean mu=3.2 and standard deviation sigma/root over n=1.1/root over 8=0.39
a. For X=3, z=(x-mu)/sigma
=(3-3.2)/0.39
=-0.51
Thus, P(X bar<3)=0.3050
b.For bar=3.5, z=(3.5-3.2)/0.39
=0.77
Thus, P(X bar>3.5)=1-P(X bar<3.5)=1-P(X bar<0.77)=1-0.7794=0.2206
c.For X=3.1, z=(3.1-3.2)/0.39=-0.25
For X bar=3.3, z=(3.3-3.2)/0.39=0.26
The required probability is: P(3.1<Xbar<3.3)
=P(X bar<3.3)-P(X bar<3.1)
=P(z<0.26)-P(z<-0.25)
=0.6026-0.4013
=0.2013