Mathematics, 02.03.2020 18:49 tay8568

An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean of 115 cm and a standard deviation of 4.7 cm. A. Find the probability that one selected subcomponent is longer than 117 cm. Probability = B. Find the probability that if 4 subcomponents are randomly selected, their mean length exceeds 117 cm. Probability = C. Find the probability that if 4 are randomly selected, all 4 have lengths that exceed 117 cm.

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Answer from: Bjehnsen3720

a) $P(X117)=P(\frac{X-\mu}{\sigma}\frac{117-\mu}{\sigma})=P(Z\frac{117-115}{4.7})=P(z0.426)$

And we can find this probability using the complement rule and we got:

P(z0.426)=1-P(z

b) $P(\bar X 117)=P(Z\frac{117-115}{\frac{4.7}{\sqrt{4}}}=0.851)$

And using the complement rule and a calculator, excel or the normal standard table we have that:

P(Z0.851)= 1-P(Z

c) $X \sim Bin (n=4, p =0.335)$

And the reason is because we assume independence between the events and each individual event have the same probability of success 0.197.

And we want this probability:

P(X=4)

We can use the probability mass function and we got:

$P(X=4) = (4C4) (0.335)^4 (1-0.335)^{4-4}= 0.0126$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

$X \sim N(115,4.7)$

Where $\mu=115$ and $\sigma=4.7$

We are interested on this probability

P(X117)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X117)=P(\frac{X-\mu}{\sigma}\frac{117-\mu}{\sigma})=P(Z\frac{117-115}{4.7})=P(z0.426)$

And we can find this probability using the complement rule and we got:

P(z0.426)=1-P(z

Part b

For this case we select a sample size of n=4, since the distribution of X is normal then the distribution for $\bar X$ is also normal and given by:

$\bar X \sim N(\mu=115, \frac{\sigma}{\sqrt{n}}= \frac{4.7}{\sqrt{4}}=2.35)$

We can find the individual probabilities like this:

$P(\bar X 117)=P(Z\frac{117-115}{\frac{4.7}{\sqrt{4}}}=0.851)$

And using the complement rule and a calculator, excel or the normal standard table we have that:

P(Z0.851)= 1-P(Z

Part c

For this case we can use a binomial distribution given by:

$X \sim Bin (n=4, p =0.335)$

And the reason is because we assume independence between the events and each individual event have the same probability of success 0.197.

And we want this probability:

P(X=4)

We can use the probability mass function and we got:

$P(X=4) = (4C4) (0.335)^4 (1-0.335)^{4-4}= 0.0126$

Answer from: Quest

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

Answer from: Quest

with btw u said 100pts but it clearly says

i answerd and just got 10pts

no hate

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