Question

It was found that the mean length of 200 diodes (LED) produced by a company

was 20.04 mm with a standard deviation of 0.02mm. Find the probability that a diode
selected at random would have a length less than 20.01mm​

Answer 1

6.68% probability that a diode selected at random would have a length less than 20.01mm​

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

`\mu = 20.04, \sigma = 0.02`

Find the probability that a diode selected at random would have a length less than 20.01mm​

This is the pvalue of Z when X = 20.01. So

`Z = (X - \mu)/(\sigma)`

`Z = (20.01 - 20.04)/(0.02)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

6.68% probability that a diode selected at random would have a length less than 20.01mm​