Respuesta :
Answer:
a) [tex]P(X>550)=P(\frac{X-\mu}{\sigma}>\frac{550-\mu}{\sigma})=P(Z>\frac{550-385}{110})=P(z>1.5)[/tex]
And we can find this probability using the complement rule and with the normal standard table or excel:
[tex]P(z>1.5)=1-P(z<1.5)=1-0.9332=0.0668 [/tex]
b) [tex]P(X<250)=P(\frac{X-\mu}{\sigma}<\frac{250-\mu}{\sigma})=P(Z<\frac{250-385}{110})=P(z<-1.227)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.227)=0.1099 [/tex]
c) [tex]P(-0.773<z<1.045)P(Z<1.045)-P(Z<-0.773)=0.8520-0.2198 =0.6322 [/tex]
d) [tex]z=1.8<\frac{a-385}{110}[/tex]
And if we solve for a we got
[tex]a=385 +1.8*110=583[/tex]
So the value of height that separates the bottom 97% of data from the top 3% is 583.
Step-by-step explanation:
Assuming this complete question: The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110.
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".
a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?
Let X the random variable that represent thecost of domestic airfares of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(385,110)[/tex]
Where [tex]\mu=385[/tex] and [tex]\sigma=110[/tex]
We are interested on this probability
[tex]P(X>550)[/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>550)=P(\frac{X-\mu}{\sigma}>\frac{550-\mu}{\sigma})=P(Z>\frac{550-385}{110})=P(z>1.5)[/tex]
And we can find this probability using the complement rule and with the normal standard table or excel:
[tex]P(z>1.5)=1-P(z<1.5)=1-0.9332=0.0668 [/tex]
b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?
[tex]P(X<250)=P(\frac{X-\mu}{\sigma}<\frac{250-\mu}{\sigma})=P(Z<\frac{250-385}{110})=P(z<-1.227)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.227)=0.1099 [/tex]
c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?
[tex]P(300<X<500)=P(\frac{300-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{500-\mu}{\sigma})=P(\frac{300-385}{110}<Z<\frac{500-385}{110})=P(-0.773<z<1.045)[/tex]
And we can find this probability with this difference:
[tex]P(-0.773<z<1.045)P(Z<1.045)-P(Z<-0.773)=0.8520-0.2198 =0.6322 [/tex]
d. What is the cost for the 3% highest domestic airfares?
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.03[/tex] (a)
[tex]P(X<a)=0.97[/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 0.97 of the area on the left and 0.03 of the area on the right it's z=1.88. On this case P(Z<1.8)=0.97 and P(z>1.8)=0.03
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.97[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.97[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.8<\frac{a-385}{110}[/tex]
And if we solve for a we got
[tex]a=385 +1.8*110=583[/tex]
So the value of height that separates the bottom 97% of data from the top 3% is 583.
