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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 centimeter. Also, suppose that the diameter of the holes in the nuts in another large box follow a normal distribution with a mean of 2.02 centimeters and a standard deviation of 0.04 centimeter. 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 centimeters. If a bolt and nut are selected at random, what is the probability they will fit together?:

My working outs:

Denote the diameter of the bolt as $d_b$ and the diameter of the holes in the nuts as $d_n$.

Then the difference is given as $d_n-d_b = 2.02-2 = 0.02$ The variance: $var(d_n-d_b)=0.03^2+0.04^2=0.0025$

Given that a bolt and a nut are selected at random, then $n=2$, and we get: $Pr\left(\frac{\sqrt{2}(d_n-d_b-0.02)}{0.0025}<0.05\right)$.

However I have a few questions on the next steps after this to clear my doubts:

Q1. Do I divide the probability of $0.05$ by $\frac{\sqrt{n}}{\sigma}$ to get $\frac{0.05}{0.0025}\sqrt{2}$? What do I do with the difference value of$-0.02$

After having a look at the solutions, I should get the form: $Pr(-0.4<Z\le0.6)$, I cannot seem to find this result - any help will be greatly appreciated!

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    $\begingroup$ Is this a homework question? $\endgroup$
    – Dave
    Commented Aug 31, 2021 at 19:05
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    $\begingroup$ @Dave No! I'm reading probability and statistics by Degroot - I have only ever self-studied , and without a lecture it's rather difficult to make out some of the terse writing $\endgroup$
    – Stackbeans
    Commented Aug 31, 2021 at 19:44
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    $\begingroup$ What do you mean by the quantity "$n$"? You contemplate one nut, one bolt, and the difference between them. There doesn't seem to be any role for an "$n$" nor for the formula you have written. $\endgroup$
    – whuber
    Commented Aug 31, 2021 at 20:35
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    $\begingroup$ $Z = (x-\mu)/\sigma$ (note the difference between standard deviation and variance) $\endgroup$
    – Dave
    Commented Aug 31, 2021 at 21:13
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    $\begingroup$ You want to find $\Pr(0 < d_n-d_b \le 0.05)$ and you know the mean and variance of $d_n-d_b$ $\endgroup$
    – Henry
    Commented Aug 31, 2021 at 23:12

1 Answer 1

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Thank you for showing your approach to this problem. I will let you resolve the sets and their probabilities.

Here is a simulation in R with a million nuts and a million bolts. The probability of a fit seems to be about $0.381 \pm 0.001.$

 set.seed(2021)
 b = rnorm(10^6, 2,.03)
 n = rnorm(10^6, 2.02,.04)
 event.1 = (n > b)
 event.2 = (abs(n-b)<.05)
 fit = event.1 & event.2
 mean(fit)
 [1] 0.381237
 2*sd(fit)/1000
 [1] 0.0009713817

In the figure below green dots are in event1 only; blue dots are in event2 only, and the red dots are in the intersecting event fit. [Apparent areas may be deceptive because the distribution of points is far from uniform.]

enter image description here

Addendum, R code for figure:

plot(n, b, pch=".")
points(n[event.1], b[event.1], pch=".", col="green")
 event.2 = (abs(n-b)<.05)
 points(n[event.2], b[event.2], pch=".", col="blue")
  fit = event.1 & event.2
  points(n[fit], b[fit], pch=".", col="red")
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  • $\begingroup$ Thanks, @Henry. Typo in copying question into code. Fortunately, most of this was still in my R session window, so the repair was not difficult. $\endgroup$
    – BruceET
    Commented Sep 2, 2021 at 5:54

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