Mr Danault - Statistics - 16 Questions, Multiple Choice & True or False

What is the probability of getting an odd number when rolling a die?
1. 1 / 6
2. 2 / 3
3. 1 / 2
4. 1 / 3
Using just the mean to describe data is okay if a manager is experienced.
1. True
2. False
Standard Deviation is a measure of data dispersion.
1. True
2. False
The median of the following data is 5: 2, 4, 2, 4, 6, 8, 4, 5, 7, 3, 11.
1. True
2. False
Joint probabilities multipled by marginal probabilities give conditional probabilities.
1. True
2. False
If events are independent then the following statement is true: p(a|b)=p(a).
1. True
2. False
The standard deviation of the following data set is 4.334: 2, 4, 2, 4, 6, 8, 4, 5, 7, 3, 11.
1. True
2. False
When a distribution is considered normal or "bell" shaped its mean and median are equal.
1. True
2. False
If p(a) = 0.5 and p(a and b) = 0.3, calculate p(b|a).
1. Not enough information
2. p(b|a) = p(a)*p(a and b) = 0.15
3. p(b|a) = p(a)/p(a and b) = 1.67
4. p(b|a) = p(a) = 0.5
5. p(b|a) = p(a and b)/p(a) = 0.6
If p(E) = 0.6, p(F) = 0.4,and p(F|E) = 0.5, calculate p(E|F):
1. p(E|F) = (p(E)*p(F|E))/p(E) = 0.5
2. Not enough information
3. p(E|F) = (p(E)*p(F|E))/p(F) = 0.75
4. p(E|F) = p(E) = 0.6
5. None of these
If p(R and B) = 0.17, p(R) = 0.6, and p(B) = 0.3, calculate p(R or B):
1. p(R or B) = p(R) + p(B) - p(R and B) = 0.73
2. p(R or B) = p(R) + p(B) = 0.9
3. No such relationships exist in probability
4. p(R or B) = p(R and B) = 0.17
5. Cannot calculate
Which represents the proper notation:
1. If event B and event J are dependent then p(B|J) = p(B).
2. If event B and event J are independent then p(B|J) = p(B).
3. If event B and event J are independent then p(B|J) = p(J|B).
4. If event B and event J are independent then p(B and J) = p(B).
5. None of these represent proper notation
Jimmy knows that 50% of the pizza slices in the fridge are stale. He also knows that given a slice of stale pizza the probability that it is plain cheese pizza is 80%. He also knows that 75% of the pizza slices in the fridge are plain cheese slices. Find the probability of Jimmy getting a stale slice of pizza given that he picks a plain cheese slice:
1. p(plain cheese|stale) = p(stale) = 0.50
2. p(plain cheese|stale) = p(stale)/p(plain cheese|stale) = 0.625
3. p(stale|plain cheese) = p(plain cheese)*p(stale) = 0.375
4. Not enough information
5. p(stale|plain cheese) = (p(stale)*p(plain cheese|stale))/p(plain cheese) = 0.533
Susie knows that if she serves her customers food that is not hot she will receive complaints. She knows that around 1% of the time the kitchen does not serve dishes hot enough (i.e., p(not hot) = 0.01). She also knows that given a customer gets a dish that is not hot that they complain 90% of the time, whereas, given a customer receives a dish that is hot enough they complain only 1% of the time. Susie would like to know what her rate of complaints will be (i.e., p(complaint)). Which of these answers is correct?
1. p(complaint) = (p(hot)*p(not hot)) /p(complaint|hot) = 0.99*0.01/0.01 = 0.99
2. p(complaint) = 0.01
3. p(complaint) = p(hot)*p(not hot) + p(complaint| not hot) = 0.99*0.01 + 0.10 = 0.1099
4. .p(complaint) = p(hot)/p(not hot) + p(complaint|hot) = 0.99/0.01 + 0.01 = 99.01
5. p(complaint) = p(hot)*p(complaint|hot) + p(not hot)*p(complaint|not hot) = 0.99*0.01 + 0.01*0.90 = 0.0999
Sabrina has a normal distribution with a mean of 75 and a standard deviation of 5. She wishes to know how approximately what percentage of values will fall between 65 and 85:
1. 10%
2. 95%
3. 68%
4. 75%
5. 99.7%
Forty people out of 100 are obese. Given that a person is obese, 80% have type II diabetes. Overall, 50 out of the 100 people have type II diabetes. Find the probability that given a person is not obese that they have type II diabetes (i.e., p(type II|not obese)).
1. p(type II|not obese) = p(type II and not obese)/p(not obese) = (p(type II)-p(obese and type II))/p(not obese) = (0.50-0.32)/0.60 = 0.30
2. p(type II|not obese) = p(type II and obese)/p(not obese) = 0.32/0.60 = 0.533
3. p(type II|not obese) = p(type II and obese)/p(obese) = 0.32/0.40 = 0.80
4. p(type II|not obese) = p(type II and not obese)/p(obese) = (0.50-0.32)/0.40 = 0.45
5. Can't tell from the data given