Respuesta :
Answer:
a) 0.0169
b) 0.19
c) 0.651
Step-by-step explanation:
Given:
Number of samples = 20
X denote the number of parts in the sample of 20 that require rework
A process problem is expected if X exceeds its mean by more than three standard deviations
Solution:
Let X be the random variable that denotes the number of parts that require rework
Then X follows a binomial distribution with parameters:
n = 20
p = 0.01
Probability mass function of X:
Pr(X=k) = [tex](\left \ {{n} \atop {k}} \right.)p^{k}(1-p)^{n-k}[/tex]
where k = 0, 1, 2, ..., 20
and
[tex](\left \ {{n} \atop {k}} \right.) = \frac{n!}{k!(n-k)!}[/tex]
a)
Compute mean and standard deviation:
percentage of parts that require rework remains at 1% so p = 0.01
n = 20
Compute Mean:
E(X) = np = 20 * 0.01 = 0.2
Compute Standard Deviation:
σ[tex]_{X}[/tex] = √np(1-p) = √20 * 0.01 ( 1 - 0.01 ) = √0.2 (0.99) = √0.198 = 0.444972
Compute upper bound:
U = E(X) + 3 * σ[tex]_{X}[/tex] = 0.2 + 3 ( 0.444972 ) = 0.2 + 1.334916 = 1.534916 = 1.53
Compute probability of X exceeds its mean by more than three standard deviations.:
P(X> 1.53 ) = 1 - P(X = 0) - P(X = 1)
P ( X = 0 ) = [tex](\left \ {{20} \atop {0}} \right.)0.01^{0}(1-0.01)^{20-0}[/tex]
= 20! / 0! ( 20 - 0 ) ! 1 ( 0.99 )²⁰
= 1 * 1 * 0.99²⁰
= 1 * 1 * 0.817907
= 0.817907
P ( X = 1 ) = [tex](\left \ {{20} \atop {1}} \right.)0.01^{1}(1-0.01)^{20-1}[/tex]
= 20! / 1! ( 20 - 1 ) ! 0.01 (0.99 )¹⁹
= 20 * 0.01 * 0.826169
= 0.165234
P(X> 1.53 ) = 1 - P(X=0) - P(X=1)= 1 - 0.817907 - 0.165234 = 0.016859 = 0.0169
b)
Compute mean and standard deviation:
rework percentage increases to 4% so p = 0.04
Compute the probability that X exceeds 1?
P(X>1) = 1 - P(X=0) - P(X=1)
P (X=0) = [tex](\left \ {{20} \atop {0}} \right.)0.04^{0}(1-0.04)^{20-0}[/tex]
= 20! / 0! ( 20 - 0 ) ! 1 ( 0.96 )²⁰
= 1 * 1 * 0.96²⁰
= 0.442002
P (X=1) = [tex](\left \ {{20} \atop {1}} \right.)0.04^{1}(1-0.04)^{20-1}[/tex]
= 20! / 1! ( 20 - 1 ) ! 0.04 (0.96 )¹⁹
= 20 * 0.04 * 0.460419
= 0.368335
P(X>1) = 1 - P(X=0) - P(X=1) = 1 - 0.442002 - 0.368335 = 0.189663 = 0.19
c)
Let Y be a random variable which denotes the number of hours in which X exceeds 1.
Since the probability that X exceeds 1 is constant in every hour, so compute p and n for Y
p[tex]_{Y}[/tex] = P(X>1)
From part (b) P(X>1) = 0.19 So,
p[tex]_{Y}[/tex] = P(X>1) = 0.19
Then Y follows a binomial distribution with parameters:
n[tex]_{Y}[/tex] = 5
p[tex]_{Y}[/tex] = 0.19
Probability mass function of Y:
Pr(Y=k) = [tex](\left \ {{n} \atop {k}} \right.)p^{k}(1-p)^{n-k}[/tex]
Pr (Y=k) = [tex](\left \ {{n} \atop {k}} \right.)[/tex] [tex]p_{k}[/tex][tex]_{Y}[/tex] (1-p[tex]_{Y}[/tex][tex])^{5-k}[/tex]
where k = 0, 1, 2, ... ,5
and
[tex](\left \ {{5} \atop {k}} \right.) = \frac{5!}{k!(5-k)!}[/tex]
Compute the probability that X exceeds 1 in at least one of the next five hours of samples:
P (Y≥1) = 1 - P (Y=0)
P(Y=0) = [tex](\left \ {{5} \atop {0}} \right.)0.19^{0}(1-0.19)^{5-0}[/tex]
= 1 * 1 * 0.81⁵
= 0.348678
P (Y≥1) = 1 - P (Y=0) = 1 - 0.348678 = 0.651322 = 0.651
