Respuesta :
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.
A similar problem is given at https://brainly.com/question/15953015
