Respuesta :
Answer:
18.15% probability of producing a bad product
Step-by-step explanation:
Problems of normally distributed samples are solved using 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 problem, we have that:
[tex]\mu = 3.01, \sigma = 0.02[/tex]
What is the probability of producing a bad product?
Less than 2.97 or more than 3.03.
Less than 2.97
pvalue of Z when X = 2.97. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.97 - 3.01}{0.02}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
More than 3.03
1 subtracted by the pvalue of Z when X = 3.03. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.03 - 3.01}{0.02}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
1 - 0.8413 = 0.1587
Then
0.0228 + 0.1587 = 0.1815
18.15% probability of producing a bad product
