Respuesta :
Answer:
a) [tex]P(X>307)=P(\frac{X-\mu}{\sigma}>\frac{307-\mu}{\sigma})=P(Z>\frac{307-268}{15})=P(z>2.6)[/tex]
And we can find this probability with the complement rule and with excel or a normal distribution table:
[tex]P(z<2.6)=1-P(z<2.6)=1-0.995=0.0045 [/tex]
b) [tex]z=-0.44<\frac{a-268}{15}[/tex]
And if we solve for a we got
[tex]a=268 -0.44*15=261.401[/tex]
So the value of length that separates the bottom 33% of data from the top 67% is 261.401.
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 lengths of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(268,15)[/tex]
Where [tex]\mu=268[/tex] and [tex]\sigma=15[/tex]
We are interested on this probability
[tex]P(X>307)[/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}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>307)=P(\frac{X-\mu}{\sigma}>\frac{307-\mu}{\sigma})=P(Z>\frac{307-268}{15})=P(z>2.6)[/tex]
And we can find this probability with the complement rule and with excel or a normal distribution table:
[tex]P(z<2.6)=1-P(z<2.6)=1-0.995=0.0045 [/tex]
Part b
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.67[/tex] (a)
[tex]P(X<a)=0.33[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.33 of the area on the left and 0.67 of the area on the right it's z=-0.44. On this case P(Z<-0.44)=0.33 and P(z>-0.44)=0.67
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.33[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.33[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.44<\frac{a-268}{15}[/tex]
And if we solve for a we got
[tex]a=268 -0.44*15=261.401[/tex]
So the value of height that separates the bottom 33% of data from the top 67% is 261.401.
