Probability question? Nuts and bolts are manufactured independently, the inner radius of nuts are normally distributed with mean 1 cm and std dev .015 and outer radius of bolts normally distriubted with mean .95 cm and std dev .015 cm. provided that the bolt radius is within .1 cm of the nut radius, the nut and bolts will fit. what is the probability of a random nut and bolt fitting?

Question

anonymous · Answer

Okay my big problem is the wording for "provided that the bolt radius is within .1 cm of the nut radius, the nut and bolts will fit"

anonymous · Answer

It turns out that if you let nut radius = x and bolt outer radius equal y it's x-y <= .1 how come?!

anonymous · Answer

any ideas silent sorrow

kropot72 · Answer

Find the multiplier for plus and minus sigma at which the difference in diameter between a nut with radius above the mean and a bolt with radius below the mean is 0.1cm as follows:
$$1+\sigma x-(0.95-\sigma x)=0.1$$
$$0.05+2\sigma x=0.1$$
$$x=\frac{0.05}{2\sigma}=1.6666.$$
Using a z-score of 1.6666 for the nut, the normal distribution table shows that the probability of a nut's radius being less than 1.6666 * sigma is 0.9521.
Can you use the normal distribution table for a z-score of -1.6666 to find the probability of a screw's radius being greater than -1.6666 * sigma?
When these two values of probability are multiplied you will have the probability of a random nut and bolt fitting.