Respuesta :
Answer:
(a) 0.94
(b) 0.20
(c) 90.53%
Step-by-step explanation:
From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let [tex]p_1[/tex] be the probability that a bearing meets the specification.
So, [tex]p_1=0.9[/tex]
Sample size, [tex]n_1=500[/tex], is large.
Let X represent the number of acceptable bearing.
Convert this to a normal distribution,
Mean: [tex]\mu_1=n_1p_1=500\times0.9=450[/tex]
Variance: [tex]\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45[/tex]
[tex]\Rightarrow \sigma_1 =\sqrt{45}=6.71[/tex]
(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.
So, [tex]X\geq 440.[/tex]
Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.
[tex]z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56[/tex].
So, the probability that a given shipment is acceptable is
[tex]P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062[/tex]
Hence, the probability that a given shipment is acceptable is 0.94.
(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.
Denote the probability od acceptance of a shipment by [tex]p_2[/tex].
[tex]p_2=0.94[/tex]
The total number of shipment, i.e sample size, [tex]n_2= 300[/tex]
Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, [tex]\mu_2[/tex], and variance, [tex]\sigma_2^2[/tex].
Mean: [tex]\mu_2=n_2p_2=300\times0.94=282[/tex]
Variance: [tex]\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92[/tex]
[tex]\Rightarrow \sigma_2=\sqrt(16.92}=4.11[/tex].
In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.
[tex]z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85[/tex].
So, the probability that a given shipment is acceptable is
[tex]P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977[/tex]
Hence, the probability that a given shipment is acceptable is 0.20.
(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).
The area right to the z-score=0.99
and the area left to the z-score is 1-0.99=0.001.
For this value, the value of z-score is -3.09 (from the z-score table)
Let, [tex]\alpha[/tex] be the required probability of acceptance of one shipment.
So,
[tex]-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}[/tex]
On solving
[tex]\alpha= 0.977896[/tex]
Again, the probability of acceptance of one shipment, [tex]\alpha[/tex], depends on the probability of meeting the thickness specification of one bearing.
For this case,
The area right to the z-score=0.97790
and the area left to the z-score is 1-0.97790=0.0221.
The value of z-score is -2.01 (from the z-score table)
Let [tex]p[/tex] be the probability that one bearing meets the specification. So
[tex]-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}[/tex]
On solving
[tex]p=0.9053[/tex]
Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.
The manufactures bearings.
As per the question, the process that a manufacturer bears 90% of the bearings and meet the thickness specifications. The shipment consists of 500 bearings. The shipment if taken has a least 440 and 500 bearings.
the answer is 0.94, 0.20, and 90.53%.
- The assumption is that each shipment is randomly taken in a sample of bearings.
- The chances or the probability that that given shipment are acceptable are 0.94% and the probability that they are more than 285 out of the 300 bearings is 0.205.
- The part of bearings that must meet the specifications in order to be 99% of shipments is 90.5%.
Learn more about the manufactures bearings.
brainly.com/question/15686369.
