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A ball bearing diameter is $3.00 \pm 0.01$.

The Mean and Standard Deviation are given (you can assume them to be any value).

Using normal distribution, find the probability of a faulty bearing.

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You most probably should explain what "faulty bearing" means. Do you mean that it is too large or too small? –  Calvin Lin Dec 30 '12 at 1:05
    
@CalvinLin any bearing having size more 3 plus minu 0.01 is declared as "Faulty Bearing". –  user54552 Dec 30 '12 at 1:09
    
Is that diameter in some units, like centimeters? –  Amzoti Dec 30 '12 at 1:16
    
@user54552 side question: Are you an engineer ? –  Amr Dec 30 '12 at 1:16
    
@Amr: sir, i am undergraduate student –  user54552 Dec 30 '12 at 6:44
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3 Answers

up vote 2 down vote accepted

Let me assume that you are new and want to understand how to set up and answer such a question. Here is a way to approach this and you should go through each step and make sure you understand!

Question

In an industrial process the diameter of a ball bearing is an important component part.

The buyer sets specifications on the diameter to be $3.00 \pm 0.01 \mbox{ cm}$.

The implication is that no part falling outside these specifications will be accepted.

It is known that in the process the diameter of a ball bearing has a normal distribution with mean $\mu = 3.0$ and standard deviation $\sigma = 0.005$.

On average, how many manufactured ball bearings will be scrapped?

Solution

The values corresponding to the specification are $x_1 = 2.99$ and $x_2 = 3.01$. Hence,

$$P(2.99 < X < 3.01) = P\left(\frac{2.99-3}{0.005} < \frac{X - 3}{0.005} < \frac{3.1-3}{0.005}\right) = P(-2.0 < Z < 2.0)$$ Using a table or calculating it, $P(Z < -2.0) = 0.0228$.

Due to symmetry of the Normal Distribution, we find that

$$\begin{align} 1 - P(-2.0 < Z < 2.0) &= 1 - P(Z < 2.0) + P(Z < -2.0) \\ &= P(Z > 2.0) + P(Z < -2.0) = 2\cdot0.0228 = 0.0456. \end{align}$$

As a result, it is anticipated that on the average, $4.56\%$ of the manufacturer's ball bearings will be scrapped.

Comment: Also review the nice form Henry used in his response and make sure that is clear.

Regards

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$\Large +1\quad\ddot\smile\quad$ –  amWhy May 10 '13 at 1:17
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I will assume the mean is $42$ and the standard deviation is $0$. Then the probability is $0$.

If the standard deviation is not zero (with a normal distribution, this suggests a positive probability of a negative diameter) then you might use $$\Phi\left(\frac{2.99-\mu}{\sigma}\right) +1- \Phi\left(\frac{3.01-\mu}{\sigma}\right)$$

where $\Phi(x)$ is the cumulative distribution function of a standard normal.

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You'd want the 68-95-99.7 rule, as a good approximation.

From what you state, I think you actually want the mean to be 3.00, and the standard deviation to be 0.01, but I may be wrong.

Under these assumptions, the probability that the R.V. drawn from a normal distribution falls within 1 standard deviation, has probability $ 1 - 2 \times(1-0.68) = 0.36$.

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