Respuesta :
Answer:
(a) 0.0169
(b) 0.189
(c) 0.651
Step-by-step explanation:
As per the question:
Total no. of samples, n = 20
Let the no. of parts requiring rework be any random variable X
The parameters for binomial distribution n = 20, p
Thus the probability distribution function in X is given by:
[tex]P(X = K) = ^{20}C_{k}(p)^{K}\times (1 - p)^{20 - K}[/tex] (1)
where
K = 0, 1, 2, ...., 20
Now,
(a) Put the value of 'p' as 0.01 in eqn (1):
Mean is given as the product of n and p as:
E(X) = [tex]n\times p = 20\times 0.01 = 0.2[/tex]
Standard deviation is given by:
[tex]\sigma = \sqrt{npq} = \sqrt{np(1 - p)} = \sqrt{20\times 0.01(1 - 0.01)} = 0.445[/tex]
If the rework is required for 2 out of 20 parts, then:
P(X > 1.53) = 1 - P(X = 0) - P(X = 1) = 0.0169
(b) When p = 0.04, probability is given by eqn (1):
P(X > 1) = 1 - P(X = 0) - P(X = 1)
[tex]P(X > 1) = 1 - (^{20}C_{0}(0.04)^{0}\times (1 - 0.04)^{20 - 0}) - (^{20}C_{1}(0.04)^{1}\times (1 - 0.04)^{20 - 1}) = 0.189[/tex]
(c) let the no. of hours (out of 5 hrs) be represented by another random variable Y, then:
[tex]p_{Y} = P(Y = K) = ^{5}C_{K}(p_{Y})^{K}\times (1 - p_{Y})^{5 - K}[/tex] (2)
Now, the probability is given by:
P[tex](Y\geq 1) = 1 - P(Y = 0) = ^{5}C_{0}(p_{Y})^{0}\times (1 - 0.189)^{5 - 0} = 0.651[/tex]
