Question

# Uninsured: It is estimated that 19% of all adults in the U.S. are uninsured. We will...

Uninsured: It is estimated that 19% 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 42 (about 14%) are uninsured.

(a) Assume the quoted value of 19% 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.

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

(c) In your survey you found 42 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.

(d) Assuming the quoted value of 19% for uninsured adults is accurate. Would 42 out of 300 be considered unusual?

Yes, that is an unusual number of uninsured adults.

No, that is not unusual.

Solution :

Given that,

p = 0.19

q = 1 - p = 1 - 0.19 = 0.81

n = 300

Using binomial distribution,

a) Mean = = n * p = 300 * 0.19 = 57

b) Standard deviation = = n * p * q = 300 * 0.19 * 0.81 = 6.8

c) x = 42

Using z-score formula,

z = x - / z = 42 - 57 / 6.8

z = -2.21

d) Yes, that is an unusual number of uninsured adults.because z-score -2.21 outside the -2.00 and 2.00 z-score

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