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

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

Your Answer:

Correct Answer:

3.2551

Comment:

P(X > x) = 0.7263 means 1 - P(X < x) = 0.7263 so P(X < x) = 1 - 0.7263 = 0.2737

Standardizing: so using the Inverse Normal we have :

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

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

## 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 =