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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

it is A...can you pls explain this??

for normal distribution,

we have to calculate z-score i.e

and then accordig to this question,P (z1 < X < z2) which is P [ ( x1 - u) / **σ < X <** (x2 - u) /**σ ]**

this is the formula..right??so what is x1 and x2 here?

if x1 = -0.005 and x2 =0.005 then P [ -0.005 - 0.9 / 0.003 < X < 0.005 - 0.9 / 0.003]..

dun know how to solve further..

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Best answer

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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