Respuesta :
Answer:
a) [tex]P(X<6) = P(Z< \frac{6-8.89}{2.11})=P(Z<-1.42)=0.0782[/tex]
b) [tex]P(8 \leq X\leq 10) = 0.3640[/tex]
c) [tex]x = 8.89 -0.842(2.11)=7.11[/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=8.89,\sigma=2.11)[/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]
a. Find P(x < 6). P(x < 6) = (Round to four decimal places as needed.)
In order to find this probability we can use the z score given by this formula
[tex]z=\frac{x-\mu}{\sigma}[/tex]
And if we use this we got this:
[tex]P(X<6) = P(Z< \frac{6-8.89}{2.11})=P(Z<-1.42)=0.0782[/tex]
b. Find P(8 < x < 10) = (Round to four decimal places as needed.)
[tex]P(8 \leq X\leq 10) = P(\frac{8-8.89}{2.11} \leq Z< \frac{10-8.89}{2.11})=P(-0.422 \leq Z\leq 0.526)=P(Z<0.526)-P(Z<-0.422)=0.701-0.337=0.3640[/tex]
c. Find the value a for which P(x < a) = 0.2. (Round to two decimal places as needed.)
For this case we can use the definition of z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
First we need to find a z score that accumulates 0.2 of the area on the left tail, and the z score on this case is z=-0.842. and using this z score we can solve for x like this:
[tex]-0.842=\frac{x-8.89}{2.11}[/tex]
And if we solve for x we got:
[tex]x = 8.89 -0.842(2.11)=7.11[/tex]
And that would be the value required for this case.
