Which statements are true? Select each correct answer. 40m6−4=4(10m6−1) 32m4+12m3=4m3(8m+3) 15m3−6m=3m(5m2−6m) 6m2+18m=6m2(1+3m)
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Answer:
(a) The value of P (X ≤ 2) is 0.8729.
(b) The value of P (X ≥ 5) is 0.0072.
(c) The value of P (1 ≤ X ≤ 4) is 0.7154.
(d) The probability that none of the 25 boards is defective is 0.2774.
(e) The expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number of defective boards.
The probability that a circuit board is defective is, <em>p</em> = 0.05.
The sample of boards selected is of size, <em>n</em> = 25.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
(a)
Compute the value of P (X ≤ 2) as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
Thus, the value of P (X ≤ 2) is 0.8729.
(b)
Compute the value of P (X ≥ 5) as follows:
P (X ≥ 5) = 1 - P (X < 5)
Thus, the value of P (X ≥ 5) is 0.0072.
(c)
Compute the value of P (1 ≤ X ≤ 4) as follows:
P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.
(d)
Compute the value of P (X = 0) as follows:
Thus, the probability that none of the 25 boards is defective is 0.2774.
(e)
Compute the expected value of <em>X</em> as follows:
Compute the standard deviation of <em>X</em> as follows:
Thus, the expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.