It is estimated that 19.4% of all adults in the U.S. are uninsured. We will assu

Question

It is estimated that 19.4% of all adults in the U.S. are uninsured. We will assume this is accurate. You take a random sample of 200 adults seen by a certain clinic and find that 49 (about 25%) are uninsured.

(a) Assume the quoted value of 19.4% for uninsured adults is accurate. What what is the mean number of uninsured adults in all random samples of size 200? Round your answer to one decimal place.
=

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

(c) In your survey you found 49 of the 200 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 = 3

(d) Assuming the quoted value of 19.4% for uninsured adults is accurate. Would 49 out of 200 be considered unusual?

Yes, that is an unusual number of uninsured adults.

No, that is not unusual.    

Explanation / Answer

this is a case of binomial approximation to normal distribution
we know that
so mean = np

here n = 200 and = 0.194
so mean = 200*0.194 = 38.8

and sd = npq , where q is 1-p
= 200*0.194*(1-0.194) = 31.27

now x = 49 and
we we know that the z score is given is as

z = (x-mean)/sd

z = (49-38.8)/31.27 = 0.3261

we see that the average value is 38.8 anf the sd is 31.27

so the value 49 would not be considered unuusual as it falls within 1 sd of the mean

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