When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05). (Round your probabilities to three decimal places.)
(a) Determine P(X ≤ 2).
(b) Determine P(X ≥ 5).
(c) Determine P(1 ≤ X ≤ 4).
(d) What is the probability that none of the 25 boards is defective?
(e) Calculate the expected value and standard deviation of X. (Round your standard deviation to two decimal places.)

expected value boards
standard deviation boards

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jennerfranco jennerfranco · Answer 1 · 2019-11-07T06:48:36+01:00

Answer:

(a) 0.873 (b) 0.007 (c) 0.715 (d) 0.277 (e) 1.25 and 1.09

Step-by-step explanation:

The probability that the random variable X takes the value x is given by P(X=x) = [tex]25Cx 0.05^x0.95^{25-x}[/tex]. Then,

(a) [tex]P(X\leq) = (25C0)(0.05^0)(0.95^{25}) + (25C1)(0.05^1)(0.95^{24}) + (25C2)(0.05^2)(0.95^{23})= 0.873[/tex]

(b) [tex]P(X\geq5) = 1-P(X\leq4) = 1 - (0.873 + (25C3)(0.05^3)(0.95^{22}) + (25C4)(0.05^4)(0.95^{21})) = 1 - (0.873 + 0.12) = 0.007[/tex]

(c) [tex]P(1\leqX\leq4) = (25C1)(0.05^1)(0.95^{24}) + (25C2)(0.05^2)(0.95^{23}) + (25C3)(0.05^3)(0.95^{22}) + (25C4)(0.05^4)(0.95^{21}) = 0.715[/tex]

(d) [tex]P(X=0) = (25C0)(0.05^0)(0.95^{25}) = 0.277[/tex]

(e) E(X) = np = (25)(0.05) = 1.25 and [tex]Sd(X) = \sqrt{Var(X)} = \sqrt{np(1-p)} = 1.09[/tex]