Respuesta :
Answer:
a) On this case [tex]\bar X \sim N(106,\frac{12}{\sqrt{25}})[/tex]
[tex]P(\bar X>110)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{110-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(Z>\frac{110-106}{\frac{12}{\sqrt{25}}})=P(Z>1.667)[/tex]
And we can find this probability on this way:
[tex]P(Z>1.667)=1-P(Z<1.667)= 1-0.952= 0.048[/tex]
b) [tex]P(106-4<\bar X<106+4)=P(\frac{102-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{110-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(\frac{102-106}{\frac{12}{\sqrt{25}}}<Z<\frac{110-106}{\frac{12}{\sqrt{25}}})=P(-1.667<Z<1.667)[/tex]
And we can find this probability on this way:
[tex]P(-1.667<X<1.667)=P(Z<1.667)-P(Z<-1.667)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.667<Z<1.667)=P(Z<1.667)-P(Z<-1.667)=0.952-0.048=0.904[/tex]
c) Figure attached
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 the variable of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(106,12)[/tex]
Where [tex]\mu=106[/tex] and [tex]\sigma=12[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
Part a
On this case [tex]\bar X \sim N(106,\frac{12}{\sqrt{25}})[/tex]
We are interested on this probability
[tex]P(\bar X>110)[/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}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X>110)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{110-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(Z>\frac{110-106}{\frac{12}{\sqrt{25}}})=P(Z>1.667)[/tex]
And we can find this probability on this way:
[tex]P(Z>1.667)=1-P(Z<1.667)= 1-0.952= 0.048[/tex]
Part b
For this case we want the probability that the sample mean lies between these two values:
[tex]P(106-4<\bar X<106+4)=P(\frac{102-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{110-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(\frac{102-106}{\frac{12}{\sqrt{25}}}<Z<\frac{110-106}{\frac{12}{\sqrt{25}}})=P(-1.667<Z<1.667)[/tex]
And we can find this probability on this way:
[tex]P(-1.667<X<1.667)=P(Z<1.667)-P(Z<-1.667)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.667<Z<1.667)=P(Z<1.667)-P(Z<-1.667)=0.952-0.048=0.904[/tex]
Part c
See the figure attached.
The probability that x-bar of the population given exceeds 110 is 0.0475.
How to compute the probability?
From the information given, the random sample of n = 25, has a mean of 106 and standard deviation of 12.
This will be converted to standard normal using z. This will be:
z = 110 - (106)/(12/✓25)
z = 1.67
P(z > 1.67) = 1 - P(z < 1.67)
= 1 - 0.9525
= 0.0475
The probability that the sample mean deviates from the population mean = 106 by no more than 4 will be calculated thus:
From the computation, Z1 and Z2 are -1.67 and 1.67 respectively.
P(102 < X < 110) = P(-1.67 < Z< 1.67)
= 0.9525 - 0.0475
= 0.9050
Learn more about probability on:
https://brainly.com/question/24756209
