INSTRUCTOR s SOLUTIONS 06/04/4 STT-35-07 SUMMER -A -04 Name MIDTERM EXAM. Given a data set 5,, 0, 3, 0, 4,, 3, 4, 4 a. 9 pts. 3+3+3 Calculate Q L, M and Q U lower quartile, median and upper quartile. M=3.5, Q L =, Q U =4. b. 4 pts. Calculate the interquartile range IQR and point out outliers, if any. Explain your conclusion on outliers. IQR=3; Q U +.5 IQR = 4+4.5 = 8.5. 0 is an upper outlier because it is to the right of 8.5. Q L -.5 IQR =.5 IQR = - 3.5. There are no lower outliers.. 5 pts. =.5+.5 Give two numerical data of size 5 such that the median of the first data is 3, and the mean is 6, while for the second data, vice versa, the median is 6 and the mean is 3. Repetitions are allowed. 3 3 3 0 - - 6 6 6
3. Consolidated Builders has bids on two large construction projects. The company president believes that the probability of winning of the first contract event A is 0.6, the probability of winning of the second contract event B is 0.5 and the probability of winning both is 0.. a. 4pts.=+ What is A B? Are the events A and B independent? Why? A B 0. A B 0.4 B 0.5 P A and B are not independent since A B A B. b. 4 pts.= + Express the following event C through A and B: C = {Consolidated wins at least one of the contracts} and find its probability. C A B C A B A B 0.6 0.5 0. 0.9. c. 4 pts. = + Express the following event D through A and B D = { Consolidated wins exactly one of the contracts} and find its probability. D A B c B A D C A B 0.9 0. 0.7. c
4. 0 pts.=5+5 Suppose 5% of the population is exposed to a television advertisement for Ford, and 3% hear advertisement on radio. Consider two events A={A randomly chosen individual is exposed to TV advertisements for Ford}; B={A randomly chosen individual is exposed to radio advertisements for Ford}. According to the text, A=.5, B=.3. Assume that events A and B are independent. Find the probability of the following events: C= {A randomly chosen individual is exposed to at least one of the advertisements}, and D={A randomly chosen individual is exposed to exactly one of the advertisements}. C A B A B D C A B.4. due to independence.5.3 0.08.49
5. A system consists of three components woring independently,connected in parallel and having the same probability of woring p. a. 7 pts. If p=0.5, what is the probability of woring of the system? - -0.5 3 = - 0.5 = 0.875. b. 8 pts. What is the value minimum of p that ensures the system reliability 0.95? p should be found from the equation: --p 3 = 0.95. Its solution gives P=0.63.
6. An economist believes that during the periods of high economic growth, the US dollar appreciates with probability 0.75, and in periods of low economic growth, the dollar appreciates with probability 0.35. The probability of high economic growth is 0.30 and the probability of low economic growth has the probability 0.70. a. 5 pts. Find the probability that for a randomly chosen moment of time the dollar appreciates. Probability of the event A A =B A B + B A B = 0.3*0.75 + 0.7*0.35 = 0.5 + 0.45= 0.470. b. 6 pts. = 3+3 Suppose the dollar appreciates at the moment. Under this condition, what are the conditional probabilities B A and B A that we are experiencing the period of high economic growth and respectively the period of low economic growth. These conditional probabilities are called posterior probabilities, in contrast to B and B which are called the prior probabilities. B B A A B A A B A A 0.5/ 0.470 0.45/ 0.470 0.479 0.5
7. Assume a random variable X has the following distribution x -5 4 6 X=x 0. 0. 0.5 0.3 a. pts. Find <X 6. Encircle the correct answer. a 0.9 b 0.6 c 0.7 d 0.8 e 0.5 b. 4 pts. Find the expected value EX of X. a 4. b 3.5 c 3.6 d 4.5 e 3.8 c. 7 pts. Find the variance VX of X. a 6.7 b.7 c 9.45 d 8.89 e.67
8. Suppose a box contains 4 red, 3 blue and 5 green balls. One ball is randomly piced 4 times with replacement. Let X be the number of green balls out of 4 piced ones. a. 3 pts. Justify the use of Binomial model for X. What are n and p? There are 4 independent trials, in each of them the green ball occurs with the probability p=0.467. X~ B 4, 0.467 b. 6 pts. Find the probability that 3 out of 4 balls are green. X=3 = 0.689.
9. Acceptance Plan. A large shipment of components will be accepted, if in a randomly chosen 00 components there are no more than defective component. Assume the defective components in the large shipment constitute 3%. a. 3 pts. Consider the random variable X = {The number of defectives in 00 chosen}. Is it a Binomial random variable? Why? What are n the number of trials and p the probability of success? There are 00 independent trials, with the probability of success p=0.03 in each of the trials. X~ B00, 0.03, n=00, p= 0.03. b. 6 pts. Find X, the probability of acceptance by use of the Binomial formula X = X=0 + X= = 0.97 00 + 00*0.03 * 0.97 99 = 0.9456. c. 8 pts. Find the same probability of acceptance X by use of the Poisson approximation X=approximately equals e, where np.! np 0003 3. X e 3 e 3 3! 4e 3 0.99.
0. Tae-home problem. a. 4 pts. Devise a problem with X~B6,0.85 Instructor s problem. The probability that a patient contracted with a disease fully recovered, is 0.85. 6 patients contracted with the disease are chosen at random. The random variable X is: The number of patients out of 6 who fully recover. b. 5 pts. Define the distribution of X. Give both the individual and cumulative probabilities. 0 3 4 5 6 0.0000 0.0004 0.0055 0.045 0.76 0.3993 0.377 0.0000 0.0004 0.0059 0.0473 0.35 0.69 c. 3 pts.=+ Determine and of X. np 5. np p 0.875. d. pts. Give the Histogram representation of the distribution. e. pts. Find X>4. X>4 = 0.35= 0.7765.
Midterm Exam Formulas The variance of a random variable is defined as follows:. :,... n n p x X V formula shortcut following the by calculated be can also X V p x p x p x p x X V The Reliability Coefficient the probability of woring for a system of components woring independently and connected in parallel equals n p P W. The Poisson Approximation to Binomial. If X~Binn,p, n is large, p is small, then.,!. p n where e appr X P