Respuesta :
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.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 20.05[/tex].
- The standard deviation is of [tex]\sigma = 0.02[/tex].
The probability that a diode selected at random would have a length less than 20.01mm is the p-value of Z when X = 20.01, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.01 - 20.05}{0.02}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] 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 https://brainly.com/question/24663213
