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

4+2=-2. 1+2=3. y=-2/3x+b. 2=4/3+b. 2/3=b. the answer is y= -2/3x+2/3

Answer from: Quest

where are the choices?

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