Respuesta :
Answer:
a) Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(12,1.25)[/tex]
Where [tex]\mu=12[/tex] and [tex]\sigma=1.25[/tex]
b) [tex]\mu_{\bar X}= 12[/tex]
c) [tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}= \frac{1.25}{\sqrt{10}}= 0.395[/tex]
d) Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu=12, \frac{\sigma}{\sqrt{n}}=0.395)[/tex]
e) [tex]P(\bar X <6)[/tex]
And we can use the z score given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this formula we got:
[tex]P(\bar X <6) = P(Z< \frac{6-12}{\frac{1.25}{\sqrt{10}}}) = P(Z<-15.17)\approx 0[/tex]
f) Yes is a problem since the probability for this event is very low so we can conclude that the event is not probable. And the sample mean is too low for the process.
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 heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(12,1.25)[/tex]
Where [tex]\mu=12[/tex] and [tex]\sigma=1.25[/tex]
Part b
For this case the mean of the sample mean [tex] \bar X[/tex] is given by:
[tex]\mu_{\bar X}= 12[/tex]
Part c
For this case the standard deviation for the sample mean would be given by:
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}= \frac{1.25}{\sqrt{10}}= 0.395[/tex]
Part d
Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu=12, \frac{\sigma}{\sqrt{n}}=0.395)[/tex]
Part e
For this case we want this probability:
[tex]P(\bar X <6)[/tex]
And we can use the z score given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this formula we got:
[tex]P(\bar X <6) = P(Z< \frac{6-12}{\frac{1.25}{\sqrt{10}}}) = P(Z<-15.17)\approx 0[/tex]
Part f
Yes is a problem since the probability for this event is very low so we can conclude that the event is not probable. And the sample mean is too low for the process.
Using the normal distribution and the central limit theorem, it is found that:
a) The random variable is the life of a dishwasher, which has [tex]\mu = 12, \sigma = 1.25[/tex]
b) The mean of the sample mean is of 12 years.
c) The standard deviation is of 0.3953 years.
d) The underlying distribution is normal, hence, the shape of the sampling distribution of the sample mean is normal.
e) 0% probability that the sample mean of the dishwashers is less than 6 years.
f) This probability is very low(Z < -2), hence, yes, it would indicate a problem with the manufacturing process.
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.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]. Also, for normal variables, this distribution is always normal, while for skewed variables, it is normal for sample sizes above 30.
Item a:
- Mean of 12 years, hence [tex]\mu = 12[/tex].
- Standard deviation of 1.25 years, hence [tex]\sigma = 1.25[/tex].
The random variable is the life of a dishwasher, which has [tex]\mu = 12, \sigma = 1.25[/tex]
Item b:
By the Central Limit Theorem, the mean of the sample mean is of 12 years.
Item c:
Sample of 10, hence [tex]n = 10[/tex] and:
[tex]s = \frac{1.25}{\sqrt{10}} = 0.3953[/tex]
The standard deviation is of 0.3953 years.
Item d:
The underlying distribution is normal, hence, the shape of the sampling distribution of the sample mean is normal.
Item e:
This probability is the p-value of Z when X = 6, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{6 - 12}{0.3953}[/tex]
[tex]Z = -15.17[/tex]
[tex]Z = -15.17[/tex] has a p-value of 0.
0% probability that the sample mean of the dishwashers is less than 6 years.
Item f:
This probability is very low(Z < -2), hence, yes, it would indicate a problem with the manufacturing process.
A similar problem is given at https://brainly.com/question/14805290
