# Multiple choice problems - stats

• Jan 11th 2009, 04:31 PM
Airjunkie
Multiple choice problems - stats
1. If P(A) =.2 and P(B)=.1 , what is P(A U B) if A and B are independent?

a) .02
b) .28
c) .30
d) .32

2. Given that 52% of the u.s. population are female and 15% of the U.S. population are older than age 65, can we conclude that (.52)(.15)=7.8% are women older than age 65?

a) Yes, by the multiplication rule.
b) Yes, by conditional probabilities.
c) Yes, by the Law of Large Number
d) No, because the events may not be independent
e) No, because the events may not be disjoint

3.Given that the probabilities P(A)=.4 and P(A U B)=.6, what is the probability P(B) if A and B are mutually exclusive (disjoint)?

a) 1
b) .6
c) .24
d) .2
e) cannot be determined

4. Which of the following pairs of events are disjoint?

a) A: the odd numbers; B: the number 5
b) A: the even numbers; B: the numbers greater than 10
c) A: the numbers less than 5; B: all negative numbers
d) A: the numbers above 100; B: the numbers less than -200
e) A: negative numbers; B: odd numbers

5. Suppose X is a random variable with a mean of m. Suppose we observe X many times and keep track of the average of the observed values. The law of large numbers says that....

a) the value of m will get larger and larger as we keep observing X.
b) this average and the value of m will both continue to get larger.
c) this average will get closer to m as we keep observing X.
d) this average will become 0.
e) this average will decrease as we keep observing X.

6. In a population of students, the number of calculators owned is a random variable X with P(X=0)=.2 , P(X=1)=.6, and P(X=2)=.2 .The mean of this probability distribution is..

a) 0
b) 2
c) 1
d) .5
e) unable to be determined

7. Refer to the information in problem #6 above. The variance of this probability distribution is..

a) 1
b) .63
c) .5
d) .4
e)unable to be determined

8) Given that 1% of batteries are ineffective, what is the probability that all 4 batteries in a package are effective.

a) .000001%
b) 1%
c) 96.06%
d) 99%
e) 99.999999%

9. The number of calories in a one-ounce serving of fruit loops is a random variable with a mean of 110. The number of calories in a cup of milk is a random variable with a mean of 140. For breakfast, you eat one ounce of cereal with 1/2 cups of milk. Let Z be the random variable that represents the total number of calories in this breakfast. The mean of Z is...

a) 110
b) 125
c) 140
d) 180
e) 250

10) The weight of a bag of candy is normally distributed with a mean of 60g and standard deviation of 2g. What is the probability that the next bag will weigh less than 55g?

a) .62%
b) 3.3%
c) 3.6%
d) 50%
e) 99.38%

-------------
1. C
2. D
3. D
4. ?
5. C
6. C
7. ?
8. C
9. E
10. B
• Jan 11th 2009, 05:09 PM
Changed 10 from b to A. Simple normal cdf(0,55,60,2)
• Jan 11th 2009, 07:56 PM
mr fantastic
Quote:

Changed 10 from b to A. Simple normal cdf(0,55,60,2)

Since it's a normal model the lower value is actually -oo.
• Jan 11th 2009, 08:05 PM
mr fantastic
Quote:

Originally Posted by Airjunkie
[snip]
4. Which of the following pairs of events are disjoint?

a) A: the odd numbers; B: the number 5
b) A: the even numbers; B: the numbers greater than 10
c) A: the numbers less than 5; B: all negative numbers
d) A: the numbers above 100; B: the numbers less than -200
e) A: negative numbers; B: odd numbers

[snip]

6. In a population of students, the number of calculators owned is a random variable X with P(X=0)=.2 , P(X=1)=.6, and P(X=2)=.2 .The mean of this probability distribution is..

a) 0
b) 2
c) 1
d) .5
e) unable to be determined

7. Refer to the information in problem #6 above. The variance of this probability distribution is..

a) 1
b) .63
c) .5
d) .4
e)unable to be determined

[snip]
-------------
1. C
2. D
3. D
4. ?
5. C
6. C
7. ?
8. C
9. E
10. B

I will help with the two you flagged with ? (Furthermore, Q6 is correct and I agree with the previous poster regarding Q10).

4. Apply the definition: Two sets are disjoint if they have no elements in common (that is, their intersection is the null set).

7. Apply the definition: $Var(X) = E(X^2) - \overline{X}^2 = E(X^2) - 1^2$ and $E(X^2)$ is readily calculated from the given data.
• Jan 11th 2009, 08:20 PM
Airjunkie
-------------
1. C
2. D
3. D
4. D new
5. C
6. C
7. ?
8. C
9. E
10. A new

Still cant grasp #7
• Jan 11th 2009, 09:13 PM
mr fantastic
Quote:

Originally Posted by Airjunkie
-------------
1. C
2. D
3. D
4. D new
5. C
6. C
7. ?
8. C
9. E
10. A new

Still cant grasp #7

Where are you stuck in calculating $E(X^2)$?