AP Statistics Unit 4

What is the probability that the person picked will be a boy given they speak german? 

• .7273

• .4

• .16

• .22

What percentage of the days that it rained, did they forecast no rain?

• About 20%

• About 18%

• About 14%

• About 12%

Match the following

• And

• Expected Value

• Binomial Distribution

• Intersection

• Geometric Distribution

• Union

• Mean

• A type of distribution that gives gives the probability that the first occurrence of success requires k independent trials.

• Or

• A type of distribution with two possible outcomes.

An event is (a)   if the probability of one variable does not affect the probability of another variable. ​ (b)  

An event is ​ (c)   if the probability of one variable indicates the other variable cannot occur. ​ ​ (d)  

• independent

• P(A)=P(A|B)

• mutually exclusive

• P(A or B) = P(A) + P(B)

Intersection (select all answers)

• P(A or B)

• P(A and B)

• P(A∩B)

• P(A∪B)

Union (select all answers)

• P(A or B)

• P(A and B)

• P(A∩B)

• P(A∪B)

States that simulated (empirical) probabilities tend to get closer to the true probability as the number of trials increases.

• 10% Rule

• Central Limit Theorem

• Law of Large Numbers

• Empirical Rule

A company states that 2% of its computers are returned for defects.  We take a sample of 100 computers to find the probability of 10 being defective.  Is this a binomial distribution?

• No because there is not a fixed number of trials.

• No because their are more than 2 outcomes.

• Yes since there are a fixed number of independent trials, outcomes in 2 categories, and the probability of defect is constant.

• No since computers would not be independent.

In the formula shown, what is in the denominator?

• Sample Mean

• Individual Score

• Sample Variance

• Sample Standard Deviation

A marketing survey compiled data on the number of personal computers in households. If X = the number of computers in a randomly-selected household, and we omit the rare cases of more than 5 computers, then X has the given distribution. The expected value of X is E(X) =

• 0.40

• 1.00

• 1.45

• 1.66

• 2.50