Respuesta :
Answer:
a) 0.2266 = 22.66% of the company's drill bits will fail before 65 hours of use.
b) 0.7734 = 77.34% will last at least 65 hours
c) 0.2266 = 22.55% will have to be replaced after more than 83 hours of use
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 = 74, \sigma = 12[/tex]
(a) What proportion of the company's drill bits will fail before 65 hours of use?
This is the pvalue of Z when X = 65.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{65 - 74}{12}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a pvalue of 0.2266
0.2266 = 22.66% of the company's drill bits will fail before 65 hours of use.
(b) What proportion will last at least 65 hours?
This is 1 subtracted by the pvalue of Z when X = 65.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{65 - 74}{12}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a pvalue of 0.2266
1 - 0.2266 = 0.7734
0.7734 = 77.34% will last at least 65 hours
(c) What proportion will have to be replaced after more than 83 hours of use?
This is 1 subtracted by the pvalue of Z when X = 65.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 74}{12}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
1 - 0.7734 = 0.2266
0.2266 = 22.55% will have to be replaced after more than 83 hours of use
