Statistics in Engineering
With examples in MATLAB^® and R

Andrew Metcalfe, David Green, Tony Greenfield, Mahayaudin Mansor, Andrew Smith and Jonathan Tuke.

Chapter 5 solutions to odd numbered exercises

Exercise 5.1

Exercise 5.3

Exercise 5.5

Exercise 5.7
Exercise 5.9

Exercise 5.11

Exercise 5.13

Exercise 5.15
Let \(X\) be the contents
1. 1. \(X \sim N(52,0.80^{2})\)
    \(P(53 \lt X) = P((53 - 52)/0.80 \lt Z)\)
    Using R
    > z=(53-52)/0.8
    > print(z)
    [1] 1.25
    > 1-pnorm(z)
    [1] 0.1056498
    > #check
    > print(1-pnorm(53,52,.8)
    + )
    [1] 0.1056498
    The proportion of bags above \(53\) is \(0.106\)
  2. Using R
    > qnorm(0.01)
    [1] -2.326348
    P(Z <-2.326)=0.01
    Z=(X-mu)/0.8
    
    So \(P((X - \mu)/0.8 \lt -2.326) = 0.01\)
    \(P(X \lt \mu - 0.8 \times 2.236) = 0.01\)
    We require
    \(\mu - 0.8 \times 2.236 = 50\)
    > mu=50+0.8*2.326
    > print(mu)
    [1] 51.8608
    The mean should be set at \(51.86\)
2. 1. Given that the bags are a random sample this corresponds to a binomial distribution with \(n = 20\) and \(p = 0.01\)
    > 1-pbinom(1,20,0.01)
    [1] 0.01685934
  2. The probability of such an extreme event, two or more underweight when the expected number under weight is \(0.2\), is rather small. Although this doesn't prove that the proportion underweight exceeds \(0.01\), it does provide convincing evidence that the proportion exceeds \(0/01\).

Exercise 5.17
1. > mu=104;sig=4.1
  > poutspec=pnorm(90,mu,sig)+(1-pnorm(110,mu,sig))
  > print(poutspec)
  [1] 0.07199632
  > pouttol=pnorm(80,mu,sig)+(1-pnorm(120,mu,sig))
  > print(1000000*pouttol)
  [1] 47.61652
  The proportion outside the internal specification is \(0.072\) and \(48\) ppm are outside the tolerance.
2. Let \(X\) be the capacitance. By symmetry, we require \(P( X \lt 90) = 0.0015\)
  > sigma=10/abs(qnorm(.0015))
  > print(sigma)
  [1] 3.36957
  The standard deviation would need to be reduced to \(3.37\)
3. > poutspec=pnorm(90,mu,sigma)+(1-pnorm(110,mu,sigma))
  > print(poutspec)
  [1] 0.03750169
  The proportion outside the internal specification would increase from \(0.003\) to \(0.038\).
  [Nearly all outside the internal spec would have too high a capacitance.
  > (1-pnorm(110,mu,sigma))
  [1] 0.03748542
  ]

Exercise 5.19
1. > L=99;U=101;mu=100.3;sig=0.8
  > pb=pnorm(L,mu,sig)
  > pa=1-pnorm(U,mu,sig)
  > cost=3+pa
  > avcost=cost/(1-pb)
  > print(avcost) [1] 3.366098
  The average cost of a within spec piston is \(3.37\)
2. 1. The mean should be set at the middle of the specification, \(100\).
  2. > L=99;U=101;mu=100;sig=0.8
    > pb=pnorm(L,mu,sig)
    > pa=1-pnorm(U,mu,sig)
    > cost=3
    > avcost=cost/(1-pb-pa)
    > print(avcost) [1] 3.803725
    The average cost of a within spec piston is \(3.80\)
    [It would be better to avoid manufacturing scrap and to try to reduce the standard deviation by improving work procedures.]

(To be continued)

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