Suppose that a certain college class contains students. Of these, are juniors, are mathematics majors, and are neither. A studen
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Answer:
And if we solve for a we got
And if we solve for a we got
So then the limits where 50% of the data lies are -47.05 and 50.35
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".
Solution to the problem
Let X the random variable that variable of interest of a population, and for this case we know the distribution for X is given by:
Where and
For this case we want the limits for the 50% of the values.
So on the tails of the distribution we need the other 50% of the data, and on ach tail we need to have 25% since the distribution is symmetric.
Lower tail
For this part we want to find a value a, such that we satisfy this condition:
(a)
(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 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 25% of data from the top 75% is -47.05.
Upper tail
For this part we want to find a value a, such that we satisfy this condition:
(a)
(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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 75% of data from the top 25% is 50.35.
So then the limits where 50% of the data lies are -47.05 and 50.35