Respuesta :
Answer:
a) [tex]\bar X \sim N(40,\frac{5}{\sqrt{64}}=0.625)[/tex]
b) [tex]P(39.6 \leq \bar X \leq 40.4)=0.4778[/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".
Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=40,\sigma=5)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
(a) What are the mean and standard deviation of the sampling distribution?
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(40,\frac{5}{\sqrt{64}}=0.625)[/tex]
(b) What is the approximate probability that x will be within 0.4 of the population mean μ? (Round your answer to four decimal places.) P =
So for this case we want this probability:
[tex]P(40-0.4 \leq \bar X \leq 40+0.4)= P(39.6 \leq \bar X \leq 40.4)[/tex]
And for this case we can use the z score given by this formula:
[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this concept we got this:
[tex]P(\frac{39.6 -40}{\frac{5}{\sqrt{64}}} \leq \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}} \leq \frac{40.4 -40}{\frac{5}{\sqrt{64}}})[/tex]
[tex]P(-0.64 \leq Z \leq 0.64) =P(z<0.64)-P(Z<-0.64)=0.7389-0.2611=0.4778[/tex]
Using the normal distribution and the central limit theorem, it is found that:
a) The mean is 40 and the standard deviation is 0.625.
b) 0.4778 = 47.78% probability that x will be within 0.4 of the population mean μ.
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].
In this problem:
- Population mean of 40, thus [tex]\mu = 40[/tex].
- Population standard deviation of 5, thus [tex]\sigma = 5[/tex]
- Sample of 64, thus [tex]n = 64[/tex].
Item a:
Bu the Central Limit Theorem:
- The mean of the sampling distribution is [tex]\mu = 40[/tex].
- The standard deviation is [tex]s = \frac{5}{\sqrt{64}} = 0.625[/tex]
Item b:
Between 39.6 and 40.4, thus, this probability is the p-value of Z when X = 40.4 subtracted by the p-value of Z when X = 39.6.
X = 40.4
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{40.4 - 40}{0.625}[/tex]
[tex]Z = 0.64[/tex]
[tex]Z = 0.64[/tex] has a p-value of 0.7389.
X = 39.6
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{39.6 - 40}{0.625}[/tex]
[tex]Z = -0.64[/tex]
[tex]Z = -0.64[/tex] has a p-value of 0.2611.
0.7389 - 0.2611 = 0.4778.
0.4778 = 47.78% probability that x will be within 0.4 of the population mean μ.
A similar problem is given at https://brainly.com/question/4202603
