Respuesta :
Answer:
a) [tex]P(X>117)=P(\frac{X-\mu}{\sigma}>\frac{117-\mu}{\sigma})=P(Z>\frac{117-115}{4.7})=P(z>0.426)[/tex]
And we can find this probability using the complement rule and we got:
[tex]P(z>0.426)=1-P(z<0.426)=1-0.665=0.335[/tex]
b) [tex]P(\bar X >117)=P(Z>\frac{117-115}{\frac{4.7}{\sqrt{4}}}=0.851)[/tex]
And using the complement rule and a calculator, excel or the normal standard table we have that:
[tex]P(Z>0.851)= 1-P(Z<0.851) =1-0.803 = 0.197[/tex]
c) [tex] X \sim Bin (n=4, p =0.335)[/tex]
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:
[tex] P(X=4)[/tex]
We can use the probability mass function and we got:
[tex] P(X=4) = (4C4) (0.335)^4 (1-0.335)^{4-4}= 0.0126[/tex]
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:
[tex]X \sim N(115,4.7)[/tex]
Where [tex]\mu=115[/tex] and [tex]\sigma=4.7[/tex]
We are interested on this probability
[tex]P(X>117)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>117)=P(\frac{X-\mu}{\sigma}>\frac{117-\mu}{\sigma})=P(Z>\frac{117-115}{4.7})=P(z>0.426)[/tex]
And we can find this probability using the complement rule and we got:
[tex]P(z>0.426)=1-P(z<0.426)=1-0.665=0.335[/tex]
Part b
For this case we select a sample size of n=4, since the distribution of X is normal then the distribution for [tex] \bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu=115, \frac{\sigma}{\sqrt{n}}= \frac{4.7}{\sqrt{4}}=2.35)[/tex]
We can find the individual probabilities like this:
[tex]P(\bar X >117)=P(Z>\frac{117-115}{\frac{4.7}{\sqrt{4}}}=0.851)[/tex]
And using the complement rule and a calculator, excel or the normal standard table we have that:
[tex]P(Z>0.851)= 1-P(Z<0.851) =1-0.803 = 0.197[/tex]
Part c
For this case we can use a binomial distribution given by:
[tex] X \sim Bin (n=4, p =0.335)[/tex]
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:
[tex] P(X=4)[/tex]
We can use the probability mass function and we got:
[tex] P(X=4) = (4C4) (0.335)^4 (1-0.335)^{4-4}= 0.0126[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 115[/tex]
- Standard deviation of 4.7 cm, hence [tex]\sigma = 4.7[/tex]
Item a:
This probability is 1 subtracted by the p-value of Z when X = 117, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{117 - 115}{4.7}[/tex]
[tex]Z = 0.426[/tex]
[tex]Z = 0.426[/tex] 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem, 4 subcomponents, hence [tex]n = 4, s = \frac{4.7}{\sqrt{4}} = 2.35[/tex]
The probability is:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{117 - 115}{2.35}[/tex]
[tex]Z = 0.85[/tex]
[tex]Z = 0.85[/tex] 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:
[tex]p = 0.3351^4 = 0.0126[/tex]
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
