Question

Uninsured: It is estimated that 17.1% of all adults in the U.S. are uninsured. We will assume this is accurate. You take a random sample of 300 adults seen by a certain clinic and find that 36 (about 12%) are uninsured.

(a) Assume the quoted value of 17.1% for uninsured adults is accurate. What what is the mean number of uninsured adults in all random samples of size 300? Round your answer to one decimal place.
μ =

(b) What is the standard deviation? Round your answer to one decimal place.
σ =

(c) In your survey you found 36 of the 300 U.S. adults are uninsured. With respect to the mean and standard deviation found in parts (a) & (b) respectively, what is the z-score for this many uninsured adults? Round your answer to two decimal places.
z =

(d) Assuming the quoted value of 17.1% for uninsured adults is accurate. Would 36 out of 300 be considered unusual?

Yes, that is an unusual number of uninsured adults.No, that is not unusual.    

Answer 1

a) = P = 0.171

b) = sqrt(P(1 - P)/n)

= sqrt(0.171(1 - 0.171)/300)

= 0.0217

c) = 36/300 = 0.12

z = ( - P)/

= (0.12 - 0.171)/0.0217

= -2.35

d) Since the z-score is less than -2, so it is an unusual value.

Yes, that is an unusual number of uninsured adults.