Respuesta :
Answer:
a) 88.54% probability of a diameter between 3.8 in and 4.3 in
b) 25.14% probability of a diameter smaller than 3.9in
c) 90.82% probability of a diameter larger than 4.2 in
Step-by-step explanation:
Problems of normally distributed samples can be 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, \sigma = 0.15[/tex]
(a) a diameter between 3.8 in and 4.3 in,
This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.
X = 4.3
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.3 - 4}{0.15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
X = 3.8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.8 - 4}{0.15}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918
0.9772 - 0.0918 = 0.8854
88.54% probability of a diameter between 3.8 in and 4.3 in
(b) a diameter smaller than 3 9 in,
This is the pvalue of Z when X = 3.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.9 - 4}{0.15}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a pvalue of 0.2514
25.14% probability of a diameter smaller than 3.9in
(c) a diameter larger than 4.2 in
This is 1 subtracted by the pvalue of Z when X = 4.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.2 - 4}{0.15}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
90.82% probability of a diameter larger than 4.2 in
