Answer:
6.68% probability that a diode selected at random would have a length less than 20.01mm
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 20.04, \sigma = 0.02[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.01 - 20.04}{0.02}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
6.68% probability that a diode selected at random would have a length less than 20.01mm
