Question

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.

Answer 1

Answer:

a) $P(X>117)=P(\frac{X-\mu}{\sigma}>\frac{117-\mu}{\sigma})=P(Z>\frac{117-115}{4.7})=P(z>0.426)$

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

$P(z>0.426)=1-P(z<0.426)=1-0.665=0.335$

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:

$P(Z>0.851)= 1-P(Z<0.851) =1-0.803 = 0.197$

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

$P(X>117)$

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:

$P(X>117)=P(\frac{X-\mu}{\sigma}>\frac{117-\mu}{\sigma})=P(Z>\frac{117-115}{4.7})=P(z>0.426)$

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

$P(z>0.426)=1-P(z<0.426)=1-0.665=0.335$

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:

$P(Z>0.851)= 1-P(Z<0.851) =1-0.803 = 0.197$

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 2

Using the normal distribution and the central limit theorem, it is found that there is a:

A. 0.3351 = 33.51% probability that one selected subcomponent is longer than 117 cm.

B. 0.1977 = 19.77% probability that if 4 subcomponents are randomly selected, their mean length exceeds 117 cm.

C. 0.0126 = 1.26% probability that if 4 are randomly selected, all 4 have lengths that exceed 117 cm.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It 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, which is the percentile of X.

In this problem:

• Mean of 115 cm, hence $\mu = 115$
• Standard deviation of 4.7 cm, hence $\sigma = 4.7$

Item a:

This probability is 1 subtracted by the p-value of Z when X = 117, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{117 - 115}{4.7}$

$Z = 0.426$

$Z = 0.426$ has a p-value of 0.6649.

1 - 0.6649 = 0.3351.

0.3351 = 33.51% probability that one selected subcomponent is longer than 117 cm.

Item b:

By the Central Limit Theorem, for sampling distributions of sample means of size n, the standard deviation is $s = \frac{\sigma}{\sqrt{n}}$.

In this problem, 4 subcomponents, hence $n = 4, s = \frac{4.7}{\sqrt{4}} = 2.35$

The probability is:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{117 - 115}{2.35}$

$Z = 0.85$

$Z = 0.85$ has a p-value of 0.8023.

1 - 0.8023 = 0.1977.

0.1977 = 19.77% probability that if 4 subcomponents are randomly selected, their mean length exceeds 117 cm.

Item c:

0.3351 = 33.51% probability that one selected subcomponent is longer than 117 cm, hence, for 4 components:

$p = 0.3351^4 = 0.0126$

0.0126 = 1.26% probability that if 4 are randomly selected, all 4 have lengths that exceed 117 cm.

A similar problem is given at https://brainly.com/question/24663213