STAT202 Study Guide - Midterm Guide: Binomial Distribution, Continuity Correction, Standard Deviation

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12 Nov 2012
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Question 1: Score 0/0
The probability of getting a parking ticket when not paying for a 2-hour period is 0.25.
What is the probability of getting at least 75 tickets if you park on 257 occasions for a
2-hour period and don’t pay?
Incorrect
Your Answer:
Correct
Answer:
0.0699
Comment:
Let X be the number of tickets received.
This is a binomial situation with p = 0.25 and n = 257,
so Mean = np = 64.25,
.
Using a Normal approximation with Continuity Correction we have
Thus, P(X ≥ 75) ≈ 1 - P(Z < 1.476579) = 0.0699.
Question 2: Score 0/0
Assume that the tires sold by Olsen Tires are normally distributed
with a mean life of 41,000 miles and a standard deviation of 2,450
miles. If you were to buy 4 Olsen tires, what is the approximate
probability that all four will last longer than 40,000 miles?
Incorrect
Your
Answer:
Correct
Answer:
0.188
Commen
t:
Let X = tire lifetime. E(X) = 41,000 and σX = 2,450. We standardize X by setting .
Now consider a single tire, and let's find the probability that its lifetime is 40,000 miles or more. That is
The probability that all four tires will last this long, assuming their lifespans are independent, is (0.6584)4 =
0.188 .
Question 3: Score 0/0
If X is a normal random variable with mean 5 and standard deviation 2.9, then find the
value x such that
P(Z > x) is equal to 0.7263, as shown below. (Note: the diagram is not necessarily to
scale.)
Incorrect
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Your Answer:
Correct Answer:
Comment:
Question 4: Score 0/0
If a baseball player's batting average is 0.314 or 31.4%, find the probability that the
player will have a bad season and only score at most 67 hits in 235 times at bat. (4
decimal accuracy).
NOTE: Please answer with a probability, not a %. For example 0.1234 instead of
12.34 .
Incorrect
Your
Answer:
Correct
Answer:
0.1883±0.001
Comment:
Let X be the number of hits in 235 at-bats. We use the normal approximation to the binomial here.
Mean
Variance so
Question 5: Score 0/0
Suppose at the University of Manitoba, 34.3% of the students live in apartments. If 178 students
are randomly selected, then the probability that the number of them living in apartments will be
between 49 and 68 inclusive, is (4 decimals):
Incorrect
Your
Answer:
Correct
0.8564±0.001
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Answer:
Comment
:
Let X be the number of students in apartments.
This is a binomial distribution with p = 0.343 and n = 178.
Mean = np = 61.054, Var = np(1 - p) = 40.1125, so SD = 6.3334
Using the normal approximation we continuity correction we have:
Question 6: Score 0/0
Defects occur in a certain manufactured tape on the average of 1 per 1,000
m. Assuming a Poisson distribution for the number of defects in a given
length of tape, what is the probability that a 4,000 m roll will have no
defects? (3 decimal accuracy)
Incorrect
Your Answer:
Correct Answer:
0.0183±0.01
Comment:
Let X = number of defects in a 4,000 m roll. X ~ Poisson(4). P(X = 0) = = 0.0183 .
Question 7: Score 0/0
A series of n independent trials are run for a Binomial Process with probability of success p. If
the mean is found to be 2.9 and the variance is 1.4, what would you estimate n to be?
Incorrect
Your Answer:
Correct Answer:
6
Comment:
We have μ = 2.9 and σ2 = 1.4 . Using the properties of the Binomial Distribution we have :
[1] np = 2.9 and
[2] np(1-p) = 1.4
Combining : 2.9(1-p) = 1.4 so or = 0.517241
Substitute this p value in [1] and solve::
For n you really should round UP to the next integer, but "normal" roundoff is accepted.
Question 8: Score 0/0
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Document Summary

The probability of getting a parking ticket when not paying for a 2-hour period is 0. 25. What is the probability of getting at least 75 tickets if you park on 257 occasions for a. Let x be the number of tickets received. This is a binomial situation with p = 0. 25 and n = 257, so mean = np = 64. 25, Using a normal approximation with continuity correction we have. Thus, p(x 75) 1 - p(z < 1. 476579) = 0. 0699. Assume that the tires sold by olsen tires are normally distributed with a mean life of 41,000 miles and a standard deviation of 2,450 miles. Now consider a single tire, and let"s find the probability that its lifetime is 40,000 miles or more. The probability that all four tires will last this long, assuming their lifespans are independent, is (0. 6584)4 =

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