Question

Uninsured Patients: It is estimated that 16.8% of all adults in the U.S. are uninsured. You take a random sample of 240 adults seen by a certain clinic and find that 47 (about 20% of them) are uninsured.

(a) Assume the 16.8% value is accurate. In all random samples of 240 U.S. adults, what is the mean and standard deviation for the number of those who are uninsured? Round both answers to 1 decimal place.

μ	=
σ	=

(b) Convert the 47 uninsured adults at this particular clinic to a z-score. Round your answer to 2 decimal places.
z =

(c) Using the normal approximation to the binomial distribution, what is the probability of getting 47 or more uninsured patients in a randomly selected group of 240 U.S. adults? Round your answer to 4 decimal places.
P(x ≥ 47) =

(d) Which of these statements describes the situation here?

It is quite possible that this higher than average proportion of uninsured patients is due to random variation.

Using the benchmark of 0.05 as a condition for something unusual, we can not definitively conclude that anything unusual is happening at this clinic.     

There is a greater than 11% chance of getting 47 or more uninsured patients in any sample of 240 patients.

All of these statements are valid.

Answer 1

Answer:

a)

Given,

sample n = 240

p = 16.8% = 0.168

Mean = np

= 240*0.168

= 40.32

Standard deviation = sqrt(npq)

= sqrt(240*0.168(1-0.168))

= 5.792

b)

Consider,

z = (x - mu)/s

substitute values

= (47 - 40.32)/5.792

z = 1.15

c)

P(x ≥ 47) = P((x-mu)/s ≥ 1.15)

= P(z ≥ 1.15)

= 0.1250719 [since from z table]

= 0.1251

d)

P(x ≥ 47) = 11%

Here we conclude that there is a greater than 11% chance of getting 47 or more uninsured patients in any sample of 240 patients.