Question

^{Suppose that the diameters of the bolts in a large box follow a normal distribution with a mean of 2 centimeters and a standard deviation of 0.03 centimeters. Also, suppose that the diameters of the holes in the nuts in another large box follow the normal distribution with a mean of 2.02 centimeters and a standard deviation of 0.04 centimeters. A bolt and a nut will fit together if the diameter of the hole in the nut is greater than the diameter of the bolt, and the difference between these diameters is not greater than 0.05 centimeter. If a bolt and a nut are selected at random, what is the probability that they will fit together?}

Short answer

The probability that a bolt and a nut will fit together is 0.3812.

Step-by-step solution

Given information

The bolts' diameter and nuts' diameter follow a normal distribution.

Calculate the probability 

Let X denote the diameter of the bolt and Y denote the diameter of the nut.

Y - X have normal distribution and

\[E\left( {Y - X} \right) = 2.02 - 2 = 0.02\]

\[V\left( {Y - X} \right) = 0.0016 + 0.0009 = 0.0025\]

If Z hasa standard normal distribution, then

\[\begin{aligned}{c}Z &= \frac{{\left( {Y - X} \right) - \mu }}{\sigma }\\ &= \frac{{\left( {Y - X} \right) - 0.02}}{{\sqrt {0.0025} }}\\ &= \frac{{\left( {Y - X} \right) - 0.02}}{{0.05}}\end{aligned}\]

Now,

\[\begin{aligned}{c}P\left( {0 < Y - X \le 0.05} \right) &= P\left( { - 0.4 < Z \le 0.06} \right)\\ &= \Phi \left( {0.06} \right) - \Phi \left( { - 0.04} \right)\\ &= \Phi \left( {0.06} \right) - 1 + \Phi \left( {0.04} \right)\\ &= 0.3812\end{aligned}\]

Therefore, the probability that a bolt and a nut will fit together is 0.3812.