8- answer is given, please show process only, thanks 5. Binomial: Airplane engines. Each engine of four (n 4) on an airplane fails11% (p = 0.11, gconditions of a binomial experiment, in other words, X is 6(4,0.11).p=0.89) of the time. Assume this problem obeys the13f(x)0.310 0.058 0.005 0.000(a) Fill in the blank: the chance no (zero) engines fail isf(0) = (O0.11°0.89* = (1) 0.005 (ii) 0.058 (iii) 0.310 (iv) 0.627.dbinom (0,4,0.11) # binomial pmf[1] 0.6274224(b) The expected number of failures isH = E(X) = f (x) = 0(0.627)+1(0.310)+2(0.058)+3(0.005)+4(0.000) =(i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73.or, using the formula, the expected number of failures isH = np = 4(0.11) =(i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73.4,1167812CC280

Solution: 5. From the given information, X follows binomial distribution with parameters n = 4 and p…

5. Binomial: Airplane engines. Each engine of four (n 11% (p = 0.11,g =1-p 0.89) of the time. Assume this problem obeys the conditions of a binomial experiment, in other words, X is b(4,0.11). 4) on an airplane fails %3D 1 3 4. | f(x) 0.310 0.058 0.005 0.000 (a) Fill in the blank: the chance no (zero) engines fail is 4 0.11°0.894 = (i) 0.005 (ii) 0.058 (iii) 0.310 (iv) 0.627. %3D %3D dbinom (0,4,0.11) # binomial pmf [1] 0.6274224 (b) The expected number of failures is H = E(X) = rf(x) = 0(0.627)+1(0.310)+2(0.058)+3(0.005)+4(0.000) = %3D %3D (i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73. or, using the formula, the expected number of failures is H=np =4(0.11) = %3D (i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73.

5. Binomial: Airplane engines. Each engine of four (n 11% (p = 0.11,g =1-p 0.89) of the time. Assume this problem obeys the conditions of a binomial experiment, in other words, X is b(4,0.11). 4) on an airplane fails %3D 1 3 4. | f(x) 0.310 0.058 0.005 0.000 (a) Fill in the blank: the chance no (zero) engines fail is 4 0.11°0.894 = (i) 0.005 (ii) 0.058 (iii) 0.310 (iv) 0.627. %3D %3D dbinom (0,4,0.11) # binomial pmf [1] 0.6274224 (b) The expected number of failures is H = E(X) = rf(x) = 0(0.627)+1(0.310)+2(0.058)+3(0.005)+4(0.000) = %3D %3D (i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73. or, using the formula, the expected number of failures is H=np =4(0.11) = %3D (i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73.

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter9: Counting And Probability

Section9.3: Binomial Probability

Problem 2E: If a binomial experiment has probability p success, then the probability of failure is...

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