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Math 365 - Sample Test No. 1
SHOW YOUR WORK. Fill in the blanks for Problems 1 – 5(a,b,c); no partial credit given for Problems 1 – 5(a,b,c). Problems 5(d,e) require a solution as well as an answer. A correct, but unsubstantiated answer on any of the Problems 5(d,e) is worth only one point. Do not use books or notes. BOX your answers. You will need your graphing calculator on some of the problems, and you are encouraged to use your calculator to verify your work. Calculator answers must be accurate to 3 decimal places.
- The age distributions of the 80 members of a statistics class are as a follows:
Age 18 19 20 21 22 23 24 Number 16 20 12 10 12 6 4
(a) Arrange the data in the following frequency table using 7 classes:
Class Frequency Relative Frequency [17.5, 18.5) [18.5, 19.5) [19.5, 20.5) [20.5, 21.5) [21.5, 22.5) [22.5, 23.5) [23.5, 24.5)
(b) Carefully sketch a frequency histogram and label the vertical axis.
18 19 20 21 22 23 24
(c) Fill in the blanks:
sample mean = sample median = sample standard deviation =
sample range = Q 1 = Q 3 =
(d) Carefully sketch a boxplot (= box and whiskers plot).
(^1718 19 20 21 22 )
- The final scores for a college statistics class are as follows:
72 62 88 58 73 84 86 60 90 82 80 70
Fill in the blanks:
(a) Sample mean x = (b) Sample standard deviation s =
(c) The interval (x–s, x+s) is ( , )
(d) Number of data points in (x–s, x+s) is.
- Three fair coins, a penny, a nickel, and a dime, are tossed and the heads and tails are recorded.
(a) List the sample space: S = { } (b) P(exactly 2 heads occur) =
(c) Find the probability that when exactly two heads occur, one of the heads is from a nickel.
P(one of the two heads is from a nickel | exactly 2 heads occur) =
- From a deck of 52 play cards, 3 cards are randomly drawn.
(a) The number of 3–card hands is.
(b) The number of 3–card hands with exactly three kings is.
(c) P(3–card hands with at least one king) =.
- (25 pts) Suppose a test for disease C is positive 95% of the time for someone with disease C and is negative 97% of the time for someone without disease C. Furthermore, suppose 1% (or 1001 ) actually have disease C. Let D = person with disease C and T = person tested positive.
(a) P(D) = P(D) =
(b) P(T|D) = P(T|D) =
(c) P(T|D) = P(T|D) =
(d) P(T∩D) = P(T∩D) =
(e) P(T) = P(D|T) =
(b) Prob =
(c) Prob =
- (a)
= 22, 100 (b)
= 4 (c)
1
2
2
1
3
0
3
- (a) P(D) =. 01 P(D) =. 99
(b) P(T|D) =. 95 P(T|D) =. 05
(c) P(T|D) =. 03 P(T|D) =. 97
(d) P(T∩D) = P(T|D)·P(D) = (.95)(.01)=. 0095 and P(T∩D) = P(T|D)·P(D) = (.03)(.99)=. 0297
(e) P(T) = P(T∩D)+P(T∩D)=.0095+.0297 =. 0392 and P(D|T) =
P(T∩D)
P(T)
