Download Test 1 - La Sierra University

Math 251, 19 April 2004, Exam 1
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Answers and Hints
Note: The references to things like RS#4 refer to question #4 on the review sheet.
1. (Similar to RS #4) (a) Systematic (b) Simple Random
2. (Similar to RS#1) (a) Ratio
(b) Nominal
(c) Cluster
(c) Interval
3. (Similar to #3,4 of Test 1, Autumn 2003) (a) Median is unchanged, the range increases by 50.
(b) Mean = (-11)(28) = -308. Standard Deviation = 77.
4. (Similar to RS#6)
Lower Upper
Lower
Upper
Cumulative Relative
Limit Limit Boundary Boundary Midpoint Frequency Frequency Frequency
10
19
9.5
19.5
14.5
4
4
.16
20
29
19.5
29.5
24.5
8
12
.32
30
39
29.5
39.5
34.5
6
18
.24
40
49
39.5
49.5
44.5
2
20
.08
50
59
49.5
59.5
54.5
5
25
.20
5. (Similar to RS#12) The median is in the 13th spot, so the median is 30; Q1= 22.5 and Q3 = 43.5
For the box plot, see the construction as in the text. The relevant numbers are: 10, 22.5, 30, 43.5, and
56 in the construction.
6. (Similar to RS#6) Construct a frequency histogram for the data using the table in (b).
Histogram
9
8
7
Frequenc y
6
5
4
3
2
1
0
.5
59
.5
49
.5
39
.5
29
.5
19
5
9.
Data
7. (Similar to RS#7) (a) 151 – 23 = 128 days
(b) 117 days
(c) 117 – 23 = 94 days
8. (Similar to RS#11) The mean is 46,500  1500 = 31
The population variance is: (1,482,000 – 46,5002/1500) 1500 = 27. Thus the population standard
deviation is 271/2  5.196
The coefficient of variation is (5.196 31)100%  16.8%
9. (Similar to RS#11(e)) (a) The interval in question is 4 standard deviations within the mean, thus at
least 15/16 of data
(b) The interval of all data within 2 standard deviations of the mean. This is 80  2(5), which gives the
interval (70,90).
10. (Similar to RS#13) Use the formulas x   xf, and x2   x2f where on the right hand side we
use the class midpoints and frequencies. Then
x 4.523 + 14.557 + 24.520 = 1420
x2  4.5223 + 14.5257 + 24.5220 = 24,455
and
SSx  24,455 – (1420)2/100 = 4291
Therefore, the mean is approximately 1420/100 = 14.2, and the sample standard deviation is
approximately s = (SSx /(n-1))1/2 = (4291/99)1/2  6.58357
11. (Similar to RS#22) (a) 9106= 9,000,000 possible phone numbers.
(b) 10009,000,000 = .000111
12. (Similar to RS#23—25)
(a) False – mutual exclusive events can be dependent
(b) True - 100/5 = 20 and add 1 to get 21
(c) False - it says at least 8/9 of data is within 3 standard deviations of the mean
(d) False - they are independent since the first draw is replaced, neither outcome affects the other
(e) False - the dot is placed above the midpoint
13. (Similar to RS#17) P(O) = 80/400 = .2, P(O and F) = 38/400 = .095, P(F) = 190/400 = .475, P(N)
= 320/400 = .80, P(N and F) = 152/400 = .38.
Alternative, one can compute P(O and F) = P(O)P(F given O) = (.2)(38/80) = .095 and P(N and F) =
P(N)P(F given N) = (.8)(152/320) = .38. You should not use the formula P(O and F) = P(O)P(F) etc,
unless you’ve already verified that O and F are independent events.
14. (Similar to RS#17) Compute P(O given F) = 38/190 = .2 and P(F given O) = 38/80 = .475.
The events O and F are independent because P(O given F) = P(O) and P(F given O) = P(F) as
computed above.
15. (Similar to RS#20) (a) C53,14 = 53! / (14! 39!) = 2.40398  1012
(b) C28,8 C25,6 = (3,108,105)(177,100) = 550,445,395,500
(c) Answer of (b)  Answer of (a) = .2290