Please I have no clue

Question 1 A physician's “patient panel” is the list of all patients under their care. In the
US, the size of patient panels among primary care doctors is well modeled by a N(1500, 200)
distribution.

(a) If a "medium-sized” panel has anywhere from 1250 to 1750 patients, what proportion
of US primary care doctors have a patient panel that is NOT "medium-sized”?
(b) Find Q1,Q3 and the IQR of the patient panels of US primary care doctors.

Answers

Answer 1

Using the normal distribution, we have that:

a) The proportion  of US primary care doctors have a patient panel that is NOT "medium-sized” is 0.2112.

b)

Q1 = 1365.

Q2 = 1635.

IQR = 270.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It measures how many standard deviations the measure is from the mean.  After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

For this problem, N(1500, 200) means that [tex]\mu = 1500, \sigma = 200[/tex].

Item a:

The proportion that are medium-sized is the p-value of Z when X = 1750 subtracted by the p-value of Z when X = 1250, thus:

X = 1750:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1750 - 1500}{200}[/tex]

[tex]Z = 1.25[/tex]

[tex]Z = 1.25[/tex] has a p-value of 0.8944.

X = 1250:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1250 - 1500}{200}[/tex]

[tex]Z = -1.25[/tex]

[tex]Z = -1.25[/tex] has a p-value of 0.1056.

0.8944 - 0.1056 = 0.7888 are medium.

1 - 0.7888 = 0.2112.

The proportion  of US primary care doctors have a patient panel that is NOT "medium-sized” is 0.2112.

Item b:

Q1 is the first quartile, which is the 25th percentile, given by X when Z has a p-value of 0.25, so X when Z = -0.675. Then:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 1500}{200}[/tex]

[tex]X - 1500 = -0.675(200)[/tex]

[tex]X = 1365[/tex]

Thus, Q1 = 1365.

Q3 is the third quartile, which is the 75th percentile, given by X when Z has a p-value of 0.75, so X when Z = 0.675. Then:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.675 = \frac{X - 1500}{200}[/tex]

[tex]X - 1500 = 0.675(200)[/tex]

[tex]X = 1635[/tex]

Thus, Q3 = 1635.

The IQR is the difference between Q3 and Q1, that is:

[tex]IQR = Q3 - Q1 = 1635 - 1365 = 270[/tex]

Then, IQR = 270.

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