Respuesta :
Answer:
a) 0.62% rings will have inside diameters exceeding 10.075 centimeters.
b) There is a 68% probability that a piston ring will have an inside diameter between 9.97 and 10.03 centimeters
c) Below 9.97cm.
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:
The finished inside diameter of a piston ring is normally distributed with a mean of 10 centimeters and a standard deviation of 0.03 centimeter. This means that [tex]\mu = 10, \sigma = 0.03[/tex]
(a) What proportion of rings will have inside diameters exceeding 10.075 centimeters ?
This proportion is 1 subtracted by the pvalue of Z when [tex]X = 10.075[/tex].
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.075 - 10}{0.03}[/tex]
[tex]Z = 2.5[/tex]
[tex]Z = 2.5[/tex] has a pvalue of 0.9938.
This means that 1-0.9938 = 0.0062 = 0.62% rings will have inside diameters exceeding 10.075 centimeters.
(b) What is the probability that a piston ring will have an inside diameter between 9.97 and 10.03 centimeters?
Those are the values that are within 1 standard deviation of the mean.
The 68-95-99.7 states that 68% of the measures of a normally distributed sample are within 1 standard deviation of the mean.
So, there is a 68% probability that a piston ring will have an inside diameter between 9.97 and 10.03 centimeters
(c) below what value of inside diameter will 15% of the piston rings fall?
This is the value of X of which Zscore has a value on the 15th percentile. This is [tex]Z = -1.035[/tex].
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.035 = \frac{X - 10}{0.03}[/tex]
[tex]X - 10 = 0.03*(-1.035)[/tex]
[tex]X = 9.97[/tex]
Below 9.97cm.
