Question

A statistical process control chart example. Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let 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.

a)If the percentage of parts that require rework remains at 1%, what is the probability that X exceeds its mean by more than three standard deviations?
b)If the rework percentage increases to 4%, what is the probability that X exceeds 1?
c)If the rework percentage increases to 4%, what is the probability that X exceeds 1 in at least one of the next five hours of samples?

Answer 1

a) 0.0169

b) 0.19

c) 0.651

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) =
`(\left \ {{n} \atop {k}} \right.)p^(k)(1-p)^(n-k)`

where k = 0, 1, 2, ..., 20

and

`(\left \ {{n} \atop {k}} \right.) = (n!)/(k!(n-k)!)`

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
`_(X)` = √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
`_(X)` = 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 ) =
`(\left \ {{20} \atop {0}} \right.)0.01^(0)(1-0.01)^(20-0)`

= 20! / 0! ( 20 - 0 ) ! 1 ( 0.99 )²⁰

= 1 * 1 * 0.99²⁰

= 1 * 1 * 0.817907

= 0.817907

P ( X = 1 ) =
`(\left \ {{20} \atop {1}} \right.)0.01^(1)(1-0.01)^(20-1)`

= 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) =
`(\left \ {{20} \atop {0}} \right.)0.04^(0)(1-0.04)^(20-0)`

= 20! / 0! ( 20 - 0 ) ! 1 ( 0.96 )²⁰

= 1 * 1 * 0.96²⁰

= 0.442002

P (X=1) =
`(\left \ {{20} \atop {1}} \right.)0.04^(1)(1-0.04)^(20-1)`

= 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
`_(Y)` = P(X>1)

From part (b) P(X>1) = 0.19 So,

p
`_(Y)` = P(X>1) = 0.19

Then Y follows a binomial distribution with parameters:

n
`_(Y)` = 5

p
`_(Y)` = 0.19

Probability mass function of Y:

Pr(Y=k) =
`(\left \ {{n} \atop {k}} \right.)p^(k)(1-p)^(n-k)`

Pr (Y=k) =
`(\left \ {{n} \atop {k}} \right.)`
`p_(k)`
`_(Y)` (1-p
`_(Y)`
`)^(5-k)`

where k = 0, 1, 2, ... ,5

and

`(\left \ {{5} \atop {k}} \right.) = (5!)/(k!(5-k)!)`

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) =
`(\left \ {{5} \atop {0}} \right.)0.19^(0)(1-0.19)^(5-0)`

= 1 * 1 * 0.81⁵

= 0.348678

P (Y≥1) = 1 - P (Y=0) = 1 - 0.348678 = 0.651322 = 0.651