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The The width of the slot of a duralumin is (in inches) normally distributed with μ=0.9 and σ=0.003. The specification limits were given as 0.9000±0.0050.
What % of forgings will be defective?

  • 0.95
  •   0.99
  •   0.98
  •  0.905
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See , unless and until we are not told it follows standard normal distribution which relies on finding Z score , we should not do so..We should follow properties of general normal distribution..

According to the question we are considering data about mean = 0.900

And value of one standard deviation                =      ∓ 0.003

But the error mentioned is                              =      ∓ 0.005

which is hence within 2 standard deviations of given data distribution..

We know ,

P(μ - σ  <=   x    <=    μ + σ)  =   0.68

P(μ - 2σ  <=   x    <=    μ + 2σ)  =   0.95

P(μ - 3σ  <=   x    <=    μ + 3σ)  =   1   , where 

μ  : Mean of normal distribution 

σ : Standard deviation of normal distribution 

For standard normal distribution ,

μ  :  0   and  σ  : 1  and   z  =  (x -  μ) /  σ

So as here it is normal distribution we need not find z score..As we can see ,

It follows the 2nd case i.e. under 2 standard deviations but not 1 standard deviation..

Hence the probabiltiy is approximately 0.95 as error limit given is 0.005 but 2 σ = 0.006..

Hence A) is the correct answer.. 

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