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.

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5. Binomial: Airplane engines. Each engine of four (n 4) on an airplane fails
11% (p = 0.11, g
conditions of a binomial experiment, in other words, X is 6(4,0.11).
p=0.89) of the time. Assume this problem obeys the
1
3
f(x)
0.310 0.058 0.005 0.000
(a) Fill in the blank: the chance no (zero) engines fail is
f(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 is
H = 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 is
H = np = 4(0.11) =
(i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73.
4,116
78
12
CC
280
Transcribed Image Text:5. Binomial: Airplane engines. Each engine of four (n 4) on an airplane fails 11% (p = 0.11, g conditions of a binomial experiment, in other words, X is 6(4,0.11). p=0.89) of the time. Assume this problem obeys the 1 3 f(x) 0.310 0.058 0.005 0.000 (a) Fill in the blank: the chance no (zero) engines fail is f(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 is H = 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 is H = np = 4(0.11) = (i) 0.44 (ii) 0.51 (iii) 0.62 (iv) 0.73. 4,116 78 12 CC 280
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