A certain diagnostic test for certain form of cancer detects the cancer if it is

Question

A certain diagnostic test for certain form of cancer detects the cancer if it is present 99% of the time. If cancer is not present it detects it incorrectly (false positive) 2% of the time. It is known from a prior study that 2.5% of the population has this type of cancer. Use a probability tree to answer the following questions: a. Find the probability that a person chosen at random from this population will test positive. b. If a person chosen at random tests positive, find the probability that the person in fact has the cancer. Repeated red blood cell counts X of the same blood sample indicate that the standard deviation is 0.025 mu where mu is the true mean count. If X is normally distributed with mean mu a. What percentage of the readings will be within 2% of the true mean mu? b. What percentage of the readings will differ from the true mean mu by at least 3%? The repair cost for a certain model of microwave oven has a normal distribution with a mean of $55, and with a standard deviation of $8. If 10 ovens are sent for repairs, find the probability that the average cost of repairs will be within $10 of the mean. The diameter of a certain species of trees has a normal distribution with mean 20 cm and standard deviation of 2 cm. A random sample of five trees was collected from a very large plantation of such trees and their diameters were measured. a. What is the probability that at all five trees selected will have diameters between 18 and 23 cm? b. What is the probability that the five trees will have an average diameter of between 18 and 23 cm? c. Your answer in part b. should be larger than your answer in part a. Explain why this must be so.

Explanation / Answer

1)a) probabilty that person test postive=really have disease and positive+not have disease & test postive

=0.025*0.99+0.975*0.02=0.04425

b)probabilty have disease given test positive =0.025*0.99/0.04425=0.559322

2)a) P(-0.02/0.025<Z<0.02/0.025)=P(-0.8<Z<0.8)=0.7881-0.2119=0.5763

b)=1-P(-0.03/0.025<Z<0.03/0.025)=1-P(-1.2<Z<1.2)=1-(0.8849-0.1151)=0.2301

3)for 10 ovens std error =std deviation/(n)1/2 =2.5298

hence P(-10/2.5298<Z<10/2.5298)=P(-3.5298<Z<3.5298)=0.999961-0.000039=0.999923

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