1.Hockey - Birthdays: It has been observed that a large percentage of profession
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Question
1.Hockey - Birthdays: It has been observed that a large percentage of professional hockey players have birthdays in the first part of the year. It has been suggested that this is due to the cut-off dates for participation in the youth leagues - those born in the earlier months are older than their peers and this advantage is amplified over the years via more opportunities to train and be coached. Of the 508 professional hockey players in a season, 157 of them were born in January, February, or March.
(a) Assume that 25% of birthdays from the general population occur in January, February, or March (these actually contain 24.7% of the days of the year). In random samples of 508 people, what is the mean number of those with a birthday in January, February, or March? Round your answer to one decimal place.
=
(b) What is the standard deviation? Round your answer to one decimal place.
=
(c) Now, 157 of the 508 professional hockey players were born in the first three months of the year. With respect to the mean and standard deviation found in parts (a) and (b) what is the z-score for 157? Round your answer to two decimal places.
z =
(d) If the men in the professional hockey league were randomly selected from the general population, would 157 players out of 508 be an unusual number of men born in the first 3 months of the year? Select One Yes or No
Yes, that is an unusual number.
No, that is not unusual.
(e) Which of the following is an acceptable sentence to explain this situation? Select one
1.If the men in the professional hockey league were selected randomly from the general population, this would be an unusual collection of birth dates.
2.There is good reason to believe that a significantly larger than expected proportion of professional hockey league players are born in the first three months of the year.
3..This could be a result of the random variation of birth dates within a sample. However, it would be pretty unlikely to happen by chance.
4.All of these are valid statements.
2. Good-Buy Electronics: You own a branch of Good-Buy Electronics and have been told by the manufacturer of Stevuski Televisions that only 5% of their brand of TVs die within one year. Last year, your branch sold 142 such televisions in July. One year later, 14 (about 10%) of them had been returned — dead.(a) Assuming the 5% value quoted by the manufacturer was accurate, what is the mean number of TVs that die within one year in randomly samples of size 142? Round your answer to one decimal place.
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(b) What is the standard deviation? Round your answer to one decimal place.
=
(c) You had 14 TVs die out of 142 (about 10%). With respect to the mean and standard deviation found in parts (a) & (b) respectively, what is the z-score for this many dead TVs? Round your answer to two decimal places.
z =
(d) Assuming the 5% rate is accurate, would 14 dead TVs in a sample of 142 TVs be considered unusual? Select One
Yes, that is an unusual number of dead TVs.
No, that is not unusual.
3. Uninsured: It is estimated that 16.8% 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 24 (about 12%) are uninsured.
(a) Assume the quoted value of 16.8% 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 24 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 16.8% for uninsured adults is accurate. Would 24 out of 200 be considered unusual? Select One Yes or NO
Yes, that is an unusual number of uninsured adults.
No, that is not unusual.
4. Cancer Survival Rates: Suppose a certain type of cancer has a 0.69 survival rate for five years. This means that 69% of those diagnosed with this type of cancer have survived at least 5 years (those that die from something else in that time are excluded). Now, suppose you check on 125 people with this type of cancer 5 years after diagnosis and treatment from a particular hospital.
(a) Assume the 0.69 survival rate is accurate. In randomly selected groups of 125 people diagnosed with this type of cancer, what is the mean number of survivors at the five-year mark? Round your answer to one decimal place.
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(b) What is the standard deviation? Round your answer to one decimal place.
=
(c) You check on 125 such patients from a certain hospital and find that only 77 survived (about 62%). With respect to the mean and standard deviation found in parts (a) & (b) respectively, what is the z-score for this many survivors? Round your answer to two decimal places.
z =
(d) While the survival rate at this particular hospital is below the national average, is 77 out of 125 low enough to be considered unusual? Select one YES or NO
Yes, that is an unusual rate.
No, it is not low enough to be considered unusual.
Explanation / Answer
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
PART A.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 508 * 0.25
= 127
PART B.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 508 * 0.25 * 0.75
= 95.25
standard deviation = sqrt( variance ) = sqrt(95.25
=9.75961
PART C.
P(X = 157) = (157-127)/9.75961
= 30/9.75961= 3.0739
PART D.
P(X > 157) = (157-127)/9.75961
= 30/9.75961 = 3.0739
= P ( Z >3.0739) From Standard Normal Table
= 0.0011
Yes, it is unusual as the result is greater than 0.05
Yes, that is an unusual number.
PART E.
4.All of these are valid statements.
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