MATE 251-A için 1

MATH 251-B 2nd Midterm December 29th, 2005

WARNİNG: Calculators, Z table and 2 pages of summaries are allowed to be used during the exam. Please type your answers between the gaps of two questions. If the gap is not enough for your answers you can use the back of the pages.

Questions

1. In a small town the given chart below reflects the ethic and religious backgrounds of the townsmen. Let C, B, I, and S denote the events of being Catholic, Baptist, Italian, and Spanish respectively.

Religion/Ethnic background	Italian	Spanish
Catholic	139	132
Baptist	114	227

Find the following:
a)    P(B)
b)    P(C\I)
c)    P(S\B)
d)    P(I\C)

Solution:

Religion/Ethnic background	Italian	Spanish	Total
Catholic	139	132	271
Baptist	114	227	341
Total	253	359	612

a) P(B)=341/612

b) P(C/I)=139/253

c) P(S/B)=227/341

d) P(I/C)=139/271

2. Suppose that the reliability of a skin test for active pulmonary tuberculosis (TB) is specified as follows: Of people with TB, 98% have a positive reaction and 2% have a negative reaction; of people free of TB, 99% have a negative reaction and 1% has a positive reaction. From a large population of which 2 per 10,000 persons have TB, a person is selected at random and given a skin test, which turns out to be positive. What is the probability that the person has active pulmonary tuberculosis?

Solution:

3. When a new machine is functioning properly, only 3% of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found

a. Describe the conditions under which the situation would be a binomial experiment.

b. what is the probability that two of the parts are defective?

c. what is the probability that none will be defective?

d. What is the probability that at least one will be a defective?

Solution:

a) This is Binomial since the out comes of the experiment have two results which is defective or not defective, selection is random and independent and we have a constant probability 0,03 which is not changing from trial to trial.

4. Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 2400 passengers per day. We assume time is Poisson Distributed

a. What is the probability that no arrivals in a 1-hour period?

b. What is the probability that no arrivals in a 15- minute period?

c. What is the probability that at least one arrival in a 15-minute period?

Solution:

a) If 2400 passengers per day than 100 passengers per hour. So our probability distribution function

is since the distribution is Poisson and expected value is 100 per hour. By this formula we can calculate the probability of no arrivals in one hour.

b) Expected value of 15 minute period is 25 since 100 passengers per hour. So our probability distribution function is by this formula we can calculate the probability of no arrivals in 25 minutes.

5. Several busses travel a certain route in Chicago so that a bus arrives at a bus stop on the route every 15 minutes. A tired shopper arrives at the bus stop at a random time in the afternoon. In this situation, it is reasonable to assume that the waiting time, x, in minutes, for the shopper until the bus arrives is uniformly distributed on the interval, [0 min, 15 min]. Find

(a) the probability the waiting time will be less than 5 minutes,

(b) the probability that x will be more than 12 minutes,

(d) the mean and standard deviation of x.

Solution:

6. At a given moment, the speeds of cars on a certain stretch of highway will be normally distributed with standard deviation 4.3 mph. It is found that 38% of the cars are traveling at speeds of 70 mph or faster. Find the mean speed of cars on this stretch of highway at any given moment.

Solution:

Second Midterm Statistics

Mean	28,5641
Standard Error	4,635979
Median	24
Mode	0
Standard Deviation	28,95168
Sample Variance	838,1997
Range	99
Minimum	0
Maximum	99
Sum	1114
Count	39