Download Solutions to sample multiple choice(1).

ELEM. STATS. MULTIPLE CHOICE SAMPLE ONE SOLUTIONS. Page 1 of 2. Assume answers
with decimals are rounded. ANSWER (e) IS ALWAYS "None of (a), (b), (c) , (d) is correct." Answer all
questions. Use 882E Scantron. Do not write on this sheet - return it & your scratch paper with your Scantron.
1.)
Which of the following statments is correct?
(a) The mean of a data set is a measure of variation.
(b) The standard deviation is a meaure of variation.
(c) The range is a measure of central tendency.
(d) The median is a measure of variation.
Mark (b) on Scantron.
2.)
Which of the following types of graphs would NOT be appropriate for quantitative data?
(a) Dot plot (b) Histogram
(c) Polygonal
(d) Pareto
Mark (d) on Scantron.
Pareto graph sorts qualitative nominal categories highest frequency to lowest.
3.)
Given the data 23, 45, 50, 70, 92, 80, 50, 31, the location of the third quartile is
(a) 70
(b) 6
(c) 6.5
(d) 75
There are 8 data. 0.75(8) = 6.
(
)
6 is a whole number, so add 0.5. L Q3 = 6.5. Mark (c) on Scantron.
4.)
Given the data 23, 45, 50, 70, 92, 80, 50, 31, the third quartile is
(a) 70
(b) 6
(c) 6.5
(d) 75 Sorted: 23, 31, 45, 50, 50, 70, 80, 92.
The mean of the 6th and 7th numbers is (70 + 80)/2 = 75.
Mark (d) on Scantron.
5.)
Given the data 23, 70, 32, 88, 50, 31, the median is (a) 32 (b) 41 (c) 60 (d) 49
n = 6. L(x ) = (n + 1 )/2 = 7/2 = 3.5. Sorted: 23, 31, 32, 50, 70, 88. The mean of the 3rd
and 4th numbers is (32 + 50)/2 = 82/2 = 41. Mark (b) on Scantron.
6.)
Find the median for the following frequency distribution:
x
2
3
4
5
f
5
10
15
30
cf
5
15
30
60
n =  f = 60.
(a) 3
(b) 3.5
(c) 4
(d) 4.5
L(x) = (n + 1 )/2 = 61/2 = 30.5. The 30th number is the last 4. The 31st number is 5.
The mean of these number is 4.5. Mark (d) on Scantron.
7.)
C(11, 6), or 11 C 6, is (a) 332,640
(b) 120 (c) 720
(d) 462
1110987
11 P 5
= 54321 = 462 Mark (d) on Scantron.
11 C 6 = 11 C 5 =
5!
8.)
A sample has mean 5.2 and standard deviation 1.5. If 4 is multiplied by each number in the
sample, the new set of data has
(a) mean 9.2, S.D. 1.5
(b) mean 9.2, S.D. 6.0
(c) mean 20.8, S.D. 6.0
(d) mean 20.8, S.D. 1.5 Multiply both mean and S.D. By 4. Mark (c) on Scantron.
9.)
Find a value of k, so that by Chebyshev's Theorem, at least 88% of the data must be
within k standard deviations of the mean. (a) k = 2.887
(b) k = 8.333 (c) k = 1.291
1
1
1
1
1
2
(d) k = 0.12 1 − 2 = .88, 1 − .88 = 2 , 0.12 = 2 , k = .12
, k =
= 2.886751346
k
Mark (a) on Scantron.
GO TO PAGE 2
k
k
.12
ELEM. STATS. MULTIPLE CHOICE SAMPLE ONE SOLUTIONS. Page 2 of 2. Assume answers
with decimals are rounded. ANSWER (e) IS ALWAYS "None of (a), (b), (c) , (d) is correct." Answer all
questions. Use 882E Scantron. Do not write on this sheet - return it & your scratch paper with your Scantron.
10.) 25% of the trees in a forest are diseased. 70% of the trees are over 10 feet.
90% are diseased OR over 10 ft. What percent of the trees are diseased AND over 10 ft?
(a) 5%
(b) 10%
(c) 15%
(d) 0%
Let D = diseased, T = over 10 ft.
(
)
(
)
(
)
(
)
P D  T = P D + P T − P D  T = .25 + .7 − .9 = .05 Mark (a) on Scantron.
11.)
12.)
60% of our parts come from supplier A, 40% from supplier B.
3% of A parts are defective, and 5% of B parts are defective.
The overall rate of defectives is (a) 3.8%
(b) 38.0%
.6(.03) + .4(.05) = .018 + .02 = .038. Mark (a) on Scantron.
(c) 4.0%
(d) 50.0%
60% of our parts come from supplier A, 40% from supplier B. 3% of A parts are defective,
and 5% of B parts are defective. The probability a defective part is from supplier B is
(a) 1/19
(b) 10/19
(c) 1/20
(d) 19/100
P(B|D) = P(B  D)/P(D) = .4(.05 )/.038 = .020/.038 = 20/38 = 10/19 Mark (b) on Scantron.
13.) Find the standard deviation for the following probability distribution:
x
1
3
4
5
p
0.3
0.4
0.1
0.2
(a) 2.09
(b) 1.4457
(c) 2.9
(d) 10.5
Enter 1, 3, 4, 5 in L1 and .3, .4, .1, .2 in L2 (without the commas.)
Run 1-Var Stats L1, L2.  x = 1.445683229
Mark (b) on Scantron.
14.) For the binomial distribution with n = 46 and p = 1/4, which values of x, the number of
successes, lie between  − 2 and  + 2 ?
(a) all
(b) 6 through 17, only (c) 9 through 14, only
(d) 11 through 12, only.
  2 = np  2 npq = 46(.25 )  2 46(.25 )(1 − .25 ) = 5.626.... & 17.37367.....
5.626.... < 6 < 17 < 17.37367.
Mark (b) on Scantron.
15.) For the binomial distribution with n = 46 and p = 1/4, find the probability of 11 successes.
(a) 0.13479
(b) 0.87175
(c) 0.13326
(d) 0.913996
11 (.75 ) 35 = binompdf(46, .25, 11 ) = .1347924728 Mark (a) on Scantron.
(
)
.25
C
46 11
16.) For the binomial distribution with n = 46 and p = 1/4, find the probability of 10 to 20
successes, inclusive. (a) 0.62146
(b) 0.74502
(c) 0.89429
(d) 0.913996
binomcdf(46, .25, 20) − binomcdf(46, .25, 9) = 0.7450224283 Mark (b) on Scantron.
17.) An urn contains 6 red marbles, 5 white and 7 blue. One at a time is drawn, without
replacement. Find the probability of a red marble on the third draw, after one white and one
blue. (a) 3/8
(b) 1/3
(c) 5/16
(d) ¼
6 red out of 16 left. 6/16 = 3/8
Mark (a) on Scantron.
ANSWERS: 1) b 2) d 3) c 4) d 5) b 6) d 7) d
12) b 13) b 14) b 15) a 16) b 17) a
GO TO PAGE 3
8) c
9) a
10) a 11) a