Respuesta :
Answer:
[tex] P(-2 \leq \bar X -\mu \leq 2)[/tex]
If we divide both sides by [tex] \frac{\sigma}{\sqrt{n}}[/tex] we got:
[tex] P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})[/tex]
And we can use the normal distribution table or excel to find the probabilites and we got:
[tex] P(-1.5 \leq Z \leq 1.5)= P(Z<1.5) -P(Z<-1.5) = 0.933-0.0668=0.866[/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".
Solution to the problem
Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(M,4)[/tex]
Where [tex]\mu=M[/tex] and [tex]\sigma=4[/tex]
We select a a sample of n =4 and 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, \frac{\sigma}{\sqrt{n}})[/tex]
And we want to find this probability:
[tex] P(-2 \leq \bar X -\mu \leq 2)[/tex]
If we divide both sides by [tex] \frac{\sigma}{\sqrt{n}}[/tex] we got:
[tex] P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})[/tex]
And we can use the normal distribution table or excel to find the probabilites and we got:
[tex] P(-1.5 \leq Z \leq 1.5)= P(Z<1.5) -P(Z<-1.5) = 0.933-0.0668=0.866[/tex]
Using the normal distribution and the central limit theorem, there is a 0.1336 = 13.36% probability that the sample mean will be within 2 square inches of the population mean.
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 standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Standard deviation of 4 square inches, hence [tex]\sigma = 4[/tex]
- Sample of 9 trees, hence [tex]n = 9, s = \frac{4}{\sqrt{9}} = 1.3333[/tex]
To find the probability that the sample mean will be within 2 square inches of the population mean, we need the following z-score:
[tex]z = \frac{2}{s}[/tex]
[tex]z = \frac{2}{1.3333}[/tex]
[tex]z = 1.5[/tex]
The probability is P(|z| < 1.5), which is 2 multiplied by the p-value of z = -1.5.
- Looking at the z-table, z = -1.5 has a p-value of 0.0668.
2 x 0.0668 = 0.1336
0.1336 = 13.36% probability that the sample mean will be within 2 square inches of the population mean.
A similar problem is given at https://brainly.com/question/24663213
