Respuesta :
Answer:
a) [tex]P(X<80)=P(\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(Z<\frac{80-72.6}{4.78})=P(z<1.548)[/tex]
And we can find this probability using the normal standard distirbution or excel and we got:
[tex]P(z<1.548)=0.939[/tex]
And that correspond to 93.9 %
b) [tex]P(60<X<80)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{60-72.6}{4.78}<Z<\frac{80-72.6}{4.78})=P(-2.636<z<1.548)[/tex]
And we can find this probability with this difference:
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator. Â
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935 [/tex]
So we have approximately 93.5%
c) [tex]z=1.64<\frac{a-72.6}{4.78}[/tex]
And if we solve for a we got
[tex]a=72.6 +1.64*4.78=80.439[/tex]
So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439. Â
d) [tex]P(X>70)=P(\frac{X-\mu}{\sigma}>\frac{70-\mu}{\sigma})=P(Z>\frac{70-72.6}{4.78})=P(z>-0.544)[/tex]
And we can find this probability using the complement rule, normal standard distirbution or excel and we got:
[tex]P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707 [/tex]
And that correspond to 70.7 %
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". Â
Part a
Let X the random variable that represent the vehicles speeds of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(72.6,4.78)[/tex] Â
Where [tex]\mu=72.6[/tex] and [tex]\sigma=4.78[/tex]
We are interested on this probability
[tex]P(X<80)[/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<80)=P(\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(Z<\frac{80-72.6}{4.78})=P(z<1.548)[/tex]
And we can find this probability using the normal standard distirbution or excel and we got:
[tex]P(z<1.548)=0.939[/tex]
And that correspond to 93.9 %
Part b
We want this probability
[tex]P(60<X<80)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{60-72.6}{4.78}<Z<\frac{80-72.6}{4.78})=P(-2.636<z<1.548)[/tex]
And we can find this probability with this difference:
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator. Â
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935 [/tex]
So we have approximately 93.5%
Part c
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.05[/tex] Â (a)
[tex]P(X<a)=0.95[/tex] Â (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a. Â
As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95[/tex] Â
[tex]P(z<\frac{a-\mu}{\sigma})=0.95[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.64<\frac{a-72.6}{4.78}[/tex]
And if we solve for a we got
[tex]a=72.6 +1.64*4.78=80.439[/tex]
So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439. Â
Part d
[tex]P(X>70)[/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>70)=P(\frac{X-\mu}{\sigma}>\frac{70-\mu}{\sigma})=P(Z>\frac{70-72.6}{4.78})=P(z>-0.544)[/tex]
And we can find this probability using the complement rule, normal standard distirbution or excel and we got:
[tex]P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707 [/tex]
And that correspond to 70.7 %
