Respuesta :
Answer:
a) There is a 15.87% probability that a part is defective.
b) The expected number of parts defective in such a sample is 0.7935.
Step-by-step explanation:
To solve this question, we use concepts of the normal probability distribution and the binomial probability distribution.
Normal probability distribution
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.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
(a) what is the probability that a part is defective?
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation
5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.
Here we have [tex]\mu = 40, \sigma = 5[/tex]
This probability is the pvalue of Z when X = 35.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 40}{5}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
There is a 15.87% probability that a part is defective.
(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in such a sample? Assume that each part is independent of the others.
This is the expected value of a binomial distribution when [tex]n = 5, p = 0.1587[/tex].
So
[tex]E(X) = np = 5*0.1587 = 0.7935[/tex]
The expected number of parts defective in such a sample is 0.7935.
