Respuesta :
Answer:
a) 0.0869 = 8.69% probability that the thickness is less than 3.0 mm
b) 0.0668 = 6.68% probability that the thickness is more than 7.0 mm
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 = 4.9, \sigma = 1.4[/tex]
(a) the thickness is less than 3.0 mm
This is the pvalue of Z when X = 3.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 4.9}{1.4}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.36[/tex] has a pvalue of 0.0869
0.0869 = 8.69% probability that the thickness is less than 3.0 mm
(b) the thickness is more than 7.0 mm
This is 1 subtracted by the pvalue of Z when X = 7. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7 - 4.9}{1.4}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% probability that the thickness is more than 7.0 mm
