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Problem set in Statistics

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Time Limit: 2hours Good luck.
1.
(25points) The following are historical data on the compressive strengths of concrete (ksi) on a certain batching plant 3.79 2.99 2.77 2.91 3.10 1.84 2.52 3.22 2.45 2.14 2.67 2.52 2.71 2.75 3.57 3.85 3.36 2.05 2.89 2.83 3.13 2.44 2.10 3.71 3.14 3.54 2.37 2.68 3.51 3.37 a.
Compute the sample mean and sample standard deviation (5points) b.
Construct a relative frequency histogram of the data (10 points) c.
Construct a steam and leaf display of the data (10 points) 2.
(45 points) Answer the following questions about probability a.
(15 points) If a multiple-choice test consists of 5 questions each with 4 possible answers of which only 1 is correct
In how many different ways can a student check off one answer to each question? (5 points)
In how many ways can a student check off one answer to each question and get all the answers wrong? (10 points) b.
(6points) The percentages of the accidents for a factory is shown below:
What is the probability that the accident occurred on the graveyard shift? (3points)
What is the probability that the accident occurred due to human error? (3points) c.
(10 points) In an experiment to study the relationship of hypertension and smoking habits, the following data are collected: Nonsmokers Moderate Smokers Heavy Smokers H 21 36 30 NH 48 26 19 where H and NH stand for Hypertension and Nonhypertension, respectively. If one of these individuals is selected at random, find the probability that the person is
Experiencing hypertension, given that the person is a heavy smoker. (5 points) Shift Unsafe Condition Human Error Day 5% 32% Evening 6% 25% Graveyard 2% 30%
a non-smoker, given that the person is experiencing no hypertension (5 ponts) d.
(9points) A regional telephone company operates three identical relay stations at different locations. During a one year period, the number of malfunctions reported by each station and the causes are shown below. Station A B C Problems with electricity supplied 2 1 1 Computer malfunction 4 3 2 Malfunctioning electrical equipment 5 4 2 Human Errors 7 7 5 Suppose that a malfunction was reported and it was found to be caused by human erros. What is the probability that it came from station C? 3.
(30 points) Answer the following probability distribution problems a.
(20 points) Three cards are drawn without replacement from the 12 face cards (jacks, queens, and kings) of an ordinary deck of 52 playing cards. Let X be the number of kings selected and Y the number of jacks. Find
The joint probability distribution of X and Y (15points)
P[(X,Y)
ε A] where A is the region given by {(x,y) , x + y ≥ 2} (5points)
b.
(10points) Let X denote the number of times a certain numerical control machine will malfunction. 1, 2, or 3 times on any given day. Let Y denote the number of times a technician is called on an emergency call. Their joint probability distribution is given as f (x,y) x 1 2 3 y 1 0.05 0.05 0.1 2 0.05 0.1 0.35 3 0 0.2 0.1
Evaluate the marginal distribution of X (8points)
Evaluate the marginal distribution of Y (7points)
Find P(Y = 3 | X = 2) (5 points)
1.
Mean = 2.90 Standard Deviation = 0.54 Stem and Leaf Stem Leaf Frequency 1 84 1 2 5, 10, 14, 37, 44, 45, 52, 52, 67, 68, 71, 75, 77, 83, 89, 91, 99 17 3 10, 13, 14, 22, 36, 37, 51, 54, 57, 71, 79, 85 12 Histogram 2.
Probability a.
n1 = 4, n2 = 4, n3 = 4, n4 = 4, n5 = 4 (n1)(n2)(n3)(n4)(n5) = 1024 n1 = 3, n2 = 3, n3 = 3, n4 = 3, n5 = 3 (n1)(n2)(n3)(n4)(n5) = 243 b.
0.02 + 0.3 = 0.32 0.32 + 0.25 + 0.3 = 0.87 c.
Consider the events A: a person is experiencing hypertension B: a person is a heavy smoker C: a person is a non-smoker
P(A/B) =30/49
P(C/A’) =48/93
012345671.82.12.42.733.33.63.9
d.
Consider the events E: a malfunction by human errors A: station A, B:station B, C:station C
||| | ||
3.
a.
If (x,y) represents the selection of x kings and y jacks in 3 draws, we must have x = 0, 1, 2, 3; y = 0, 1, 2, 3; and 0
≤ x + y ≤
3. Therefore (2,1) represents the selection of 2 kings and 1 jack, which will occur with probability
(
)(
)(
)
After finding all the probabilities for other possibilities, we construct the JPD. f (x,y) x 0 1 2 3 y 0 1/55 6/55 6/55 1/55 1 6/55 16/55 6/55 2 6/55 6/55 3 1/55
P[(X,Y)
ε A]
= P(x + y ≥ 2) = 1 –
P(x + y < 2) = 1
–
1/55
–
6/55
–
6/55 = 42/55 b.
Marginal Distribution
Marginal Distribution of X x 1 2 3 g(x) 0.05 + 0.05 + 0 0.05 + 0.1 + 0.2 0.1 + 0.35 +0.1
Marginal Distribution of Y x 1 2 3 h(y) 0.05 + 0.05 + 0.1 0.05 + 0.1 + 0.2 0 + 0.2 + 0.1
P(Y = 3 | X = 2) = 0.2/(0.05 + 0.1 + 0.2) = 0.57

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