Uninsured: It is estimated that 17.2% of all adults in the U.S. are uninsured. W
ID: 3195657 • Letter: U
Question
Uninsured: It is estimated that 17.2% 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 44 (about 22%) are uninsured.
(a) Assume the quoted value of 17.2% 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 44 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 =
(d) Assuming the quoted value of 17.2% for uninsured adults is accurate. Would 44 out of 200 be considered unusual?
Yes, that is an unusual number of uninsured adults.No, that is not unusual.
Explanation / Answer
(a) Assume the quoted value of 17.2% for uninsured adults is accurate. What what is the mean number of uninsured adults in all random samples of size 200?
Answer : Mean number of uninsured adults = 200 * 0.172 = 34.4
(b) What is the standard deviation?
Answer : standard deviation = sqrt [200 * 0.172 * (1 - 0.172)] = 5.337
(c) In your survey you found 44 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?
Answer : Z = (44 - 34.3)/5.337 = 1.8175 or 1.82
(d) Assuming the quoted value of 17.2% for uninsured adults is accurate. Would 44 out of 200 be considered unusual?
No, that is not unusual as Z values is less than 2 , so, it is not unusual.
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