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.
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.
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$.
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.
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