Using the normal distribution, it is found that there is a 0.0228 = 2.28% probability that a diode selected at random would have a length less than 20.01mm.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem, we have that:

The mean is of $\mu = 20.05$ .

The standard deviation is of $\sigma = 0.02$ .

The probability that a diode selected at random would have a length less than 20.01mm is the <u>p-value of Z when X = 20.01</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.01 - 20.05}{0.02}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228.

0.0228 = 2.28% probability that a diode selected at random would have a length less than 20.01mm.

More can be learned about the normal distribution at brainly.com/question/24663213

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