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Unformatted text preview: STAT 350 – Spring 2009
Exam 1 SOLUTION
Your Name: ____________________________________ Your Seat: __________
Section Time (circle): 10:30 11:30 4:30 Note:
• You are responsible for upholding the Honor Code of Purdue University. This includes
protecting your work from other students.
• Show your work on all questions. Unsupported work will not receive full credit. Credit
will not be given for dumb luck. Showing work includes defining any random variable or
event you use in the solution. Showing work also includes identifying any named
distribution you are using in the solution and the values of all relevant parameters.
• Make it clear what your final answer is to each question. It may help to circle your
answer.
• Decimal answers should be exact or to at least four significant digits (exceptions as
discussed in class).
• Standard Normal (Z) values/probabilities must be taken from the table provided.
• You are allowed the following aids: a onepage (8.5×11 inch) cheat sheet, a scientific
calculator, and pencils.
• Turn off and put away your cell phone before the exam begins! Question Points Possible Points Missed
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10
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100 Exam Score: _______________/ 100 1. (10 points; 5 points each part) Use the following set of number to answer the questions
below. Assume you will be making a histogram of this data using the following cutpoints:
0, 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100.
3.695
26.671
43.833
52.710
59.652
70.249
78.219
82.719
87.207
92.490 14.007
28.106
43.981
55.004
61.486
70.682
78.477
83.089
88.380
93.680 17.703
28.805
44.943
55.620
63.623
71.429
78.900
83.852
89.003
93.881 18.947
31.591
45.430
55.669
65.098
72.494
79.272
84.843
89.890
95.002 20.453
33.202
48.031
55.723
65.722
72.660
80.005
85.069
90.252
95.864 20.692
33.939
49.455
56.312
67.247
73.517
81.201
85.271
90.469
97.761 20.723
35.026
49.755
56.748
67.647
75.063
81.700
86.262
91.064
97.811 21.507
39.994
52.470
58.890
70.169
76.971
82.501
87.021
91.169
97.952 a. If you are making a frequency histogram, what would be the height of the bar for the 30
to 40 bin?
5
There are 5 observations (highlighted) with values between 30 and 40 b. If you are making a density histogram, what would be the height of the bar for the 30 to
40 bin?
Relative Frequency = (# of observation in that class) / (total # of observations)
Density = (relative frequency) / (class width) = (5/80)/10 = 0.00625 2. (10 points; 5 points each part) Base your answers to the questions below on the following
data set:
21 33 38 44 47 54 54 55 60 61 66 67 68 69 70 71 72 76 93 a. Find the median of this data set.
61 b. Find the lower quartile of this data set.
The lower quartile is the median of the smallest 10 observations (highlighted)
LQ is halfway between 47 and 54: (47 + 54) / 2 = 50.5 STAT 350 Exam 1 Page 2 of 2 3. (10 points; 5 points each part) You have a sample of 35 observations from a population, you
35 find that 35 ∑ x = 2288 and ∑ x
i =1 i i =1 2
i = 174,526. a. Find x
2288 / 35 = 65.37143 b. Find sx 2
sx = SS xx
=
n −1 ( 2288)
174526 −
35 − 1 35 2 = 24956.17143
= 734.005042
34 2
sx = sx = 27.09252742 4. (10 points) X is a random variable with the following PDF. Find the mean of X.
⎧0.34 − 0.04 x if 1 < x < 6
f ( x) = ⎨
0
otherwise
⎩ μ= +∞ ∫ xf ( x ) dx −∞ STAT 350 Exam 1 6 = ∫ x ( 0.34 − 0.04 x ) dx = 37/12 = 3.08333 1 Page 3 of 3 5. (10 points) For Spring Break, 48% of Purdue undergraduates will go to Florida, 22% will go
to Mexico and the rest will stay in Indiana. 40% of the undergraduates that go to Florida will
get sunburns, 30% of the undergraduates that go to Mexico will get sunburns, and only 1% of
the undergraduates that stay in Indiana will get sunburns. What percent of all Purdue
undergraduates will get sunburns over Spring Break?
P(Burn)
= P(Florida)P(Burn  Florida) + P(Mexico)P(Burn  Mexico) + P(Indiana)P(Burn Indiana)
= (0.48)(0.40) + (0.22)(0.30) + (0.30)(0.01) = 0.261
26.1% 6. (10 points) You randomly select 27 tulip bulbs from a large discount bin at your local
nursery. The nursery manager tells you that 84% of the bulbs in the bin are for red tulips. If
you plants these bulbs and all 27 bloom, what is the probability you will get at least 25 red
tulips?
X = # of red tulips
X ~ Binomial (n = 27, pi = 0.84)
n(1 – pi) < 5, therefore the normal approximation to the binomial is NOT appropriate
P(X ≥ 25) = P(X = 25) + P(X = 26) + P(X = 27)
⎛ 27 ⎞ 25 2 ⎛ 27 ⎞ 26 1 ⎛ 27 ⎞ 27 0
⎜ ⎟ .84 .16 + ⎜ ⎟ .84 .16 + ⎜ ⎟ .84 .16
⎝ 25 ⎠
⎝ 26 ⎠
⎝ 27 ⎠
= 0.1704066 STAT 350 Exam 1 Page 4 of 4 7. (10 points; 5 points each part) In the Hoosier Lottery, if you picked 2 of the 6 winning
numbers, you win a free “quick pick”. The probability of winning a free quick pick is 0.137.
Assume you play the Hoosier Lottery once a week for the next 15 weeks.
a. What is the expected value of the proportion of weeks in which you win a free quick
pick?
E(p) = π = 0.137 b. What is the standard deviation of the proportion of weeks in which you win a free quick
pick? SD ( p ) = π (1 − π )
n = 0.137 (1 − 0.137 )
= 0.088781
15 8. (10 points) The lifetime of a particular type of battery has a mean of 19.4 hours with a
standard deviation of 4.1 hours. If you have 50 of these batteries, what is the probability that
the average lifetime of those 50 batteries is between 18 and 19 hours?
E(X1) = 19.4
SD(X1) = 4.1
n = 50
4.1 ⎞
⎛
X ~ Normal ⎜ μ = 19.4, σ =
⎟
50 ⎠
⎝
⎛ 18 − 19.4
19 − 19.4 ⎞
P (18 < X < 19 ) = P ⎜
<Z <
⎟ = P(−2.41 < Z < −0.69)
4.1 50 ⎠
⎝ 4.1 50
= Φ ( −0.69 ) − Φ ( −2.41) = 0.2451 − 0.0080
= 0.2371 STAT 350 Exam 1 Page 5 of 5 9. (10 points) Bodymassindex (BMI) is a ratio of weight to height or length. In humans, high
values of BMI indicate a person is overweight (too heavy for their height). In wildlife, BMI
(usually measured as a ratio of weight to length) is used as a measure of “condition” individuals with high BMIs are considered healthy and those with low BMIs are considered
to be in underweight and, therefore, in poor health (obesity doesn’t happen in the wild). 35
black bears from the George Washington and Jefferson National Forest in Virginia were
captured and the BMI was calculated for each bear. For the 35 bears sampled, the average
BMI was 246 and the standard deviation was 41. Obtain an 80% confidence interval for the
true mean BMI of the population of black bears living in the George Washington and
Jefferson National Forest in Virginia.
α = 1 – 0.80 = 0.20
zα/2 = Φ1(1 – 0.20/2) = Φ1(0.90) = 1.28
80% CI: 246 ± 1.28 41
→
35 (237.12926, 254.8707) 10. (10 points) You wish to estimate the proportion of black bears in the National Forests that
are underweight (have a BMI below a particular value). How many bears will you need to
sample to ensure that your resulting 95% CI will be within 0.05 of the true value?
α = 1 – 0.95 = 0.05
zα/2 = Φ1(1 – 0.05/2) = Φ1(0.975) = 1.96
⎡z ⎤
n = π (1 − π ) ⎢ α /2 ⎥
⎣B⎦ 2 ⎡ 1.96 ⎤
= 0.5 (1 − 0.5 ) ⎢
⎣ 0.05 ⎥
⎦ 2 = 384.16 2 ⎡z ⎤
384 or 385 are also a π (1 − π ) ⎢ α /2 ⎥ acceptable answers
⎣B⎦ STAT 350 Exam 1 Page 6 of 6 ...
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This note was uploaded on 02/16/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Staff

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