The median survival time the average the and ordered survival times


Solution: Valid systolic blood pressure values exist on a continuum. Between any two values, it is theoretically possible for there to exist an intermediate value. Division of two systolic blood pressure values produces a meaningful number (not clinically relevant but well defined anyway!)

b. (2 points)
Construct a figure that displays simultaneously the cumulative relative frequency distribution for nonsmokers and smokers.

Practicetest1_solutions.doc

25-16 =	P 25	-109	⇒	P =1 13. 7
54-16	129-109

75-54 =	P 75	-129	⇒
	149-129			75

Note – If your solution used the midpoints of the intervals to obtain guesses of the 25th and 75th percentiles, I will give you full credit - cb.

50-16 =	P 50	-109	⇒
54-16	129-109			50

Construct a box plot for systolic blood pressure among non-smokers.

4. (10 points)

a.


Solution: A person cannot be simultaneously type A or type AB blood type; thus these events are mutually exclusive and so the probability of their union is the simple sum. Pr [ Type “A” or “AB” ] = Pr [Type “A”] = Pr [ Type “AB” ] = 0.3780 + 0.0372 = 0.4152

d.

Answer: New mean = 45+10=55 New median = 21 + 10 = 31 New standard deviation = 15 (no change) New sample variance = 225 (no change) New interquartile range = 32 (no change)

Practicetest1_solutions.doc

1.

2.

Sample standard deviation
= S 2
= 1.0875
=1.0428

Practicetest1_solutions.doc

b.

Answer: 88.89%


Expected # positive = [#disease] Pr[+\|disease] + [#without disease] Pr[+\|no disease] = [1](.9167) + [999,999](1-.8889) = [.9167] + [111,099.8889] = 111,100.8056 Expected proportion postive = [111,100.8056]/[1,000,000] = .1111 = 11.11%


Solution: Expected # false positive = [#without disease] Pr[+\|no disease] = [999,999](1-.8889) = 111,099.8889 ≈ 111,100


Solution: The most frequently occurring survival time is the modal class. This is the interval “less than one year”.

b.

Answer: No.
Solution: Not available are the actual survival times of the 47 persons whose survival times are at least 20 years.

9.

Answer: 25
Solution: Again, as distribution is symmetric, average is center value or 25, approximately

Answer: Statement (i)
Solution: Undergraduate classes are more likely to include some number of adult learners than precocious teen learners. The result is positive skewness.