Describing Probability Distributions

A marketing survey compiled data on the total number of televisions in households. If X = the number of televisions in a randomly-selected household, and we omit the rare cases of more than 5 televisions, then X has the distribution below. What is the probability that a randomly chosen household has at least two televisions?

• 0.19

• 0.20

• 0.29

• 0.39

• 0.61

Tags

CCSS.HSS.MD.A.3

CCSS.HSS.MD.A.4

A random variable X has a probability distribution as follows Where k is a positive constant. The probability P(X < 2.0) is equal to

• 0.90

• 0.25

• 0.65

• 0.15

• 1.00

Tags

CCSS.HSS.MD.A.3

CCSS.HSS.MD.A.4

A business evaluates a proposed venture as follows. It stands to make a profit of $10,000 with probability 3/20, to make a profit of $5000 with probability 9/20, to break even with probability 5/20, and to lose $5000 with probability 3/20. The expected profit in dollars is

• 1500

• 0

• 3000

• 3250

• -1500

Tags

CCSS.HSS.MD.A.2

Roll one 8-sided die 10 times. The probability of getting exactly 3 sevens in those 10 rolls is given by

• 
• 
• 
• 
• 

Tags

CCSS.HSS.MD.B.6

CCSS.HSS.MD.B.7

X and Y are independent random variables, and a and b are constants. Which one of the following statements is true?

• $\sigma_{X+Y}\ =\ \sigma_X+\sigma_Y$

• Var (X-Y) = Var (X) + Var (Y)

• Var(a+bX)=bVar(X)

• $\sigma_{X+Y}=\sigma_X-\sigma_Y$

• $Var\ \left(X+Y\right)\ =\ \sqrt{Var\left(X^2\right)+Var\left(Y^2\right)}$

Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume that X is Normal with mean $360 and standard deviation $50. What is P(X > $400)?

• 0.2119

• 0.2881

• 0.5319

• 0.7881

• 0.8450

Tags

CCSS.HSS.ID.A.4

You select 40 cards from a standard deck of 52 cards. Let Y be the number of red cards (hearts or diamonds) in the 40 cards selected. Which of the following best describes this setting?

• Y has a binomial distribution with n = 40 observations and probability of success p = 0.5.

• Y has a binomial distribution with n = 40 observations and probability of success p = 0.5, provided the deck is shuffled well.

• Y has a binomial distribution with n = 40 observations and probability of success p = 0.5, provided that after selecting a card it is replaced in the deck and the deck is shuffled well before the next card is selected.

• Y has a geometric distribution with n = 40 observations and probability of success p = 0.5.

• Y has a geometric distribution with n = 52 observations and probability of success p = 0.5.

Tags

CCSS.HSS.MD.A.3

CCSS.HSS.MD.A.4

In a large population of college students, 20% of the students have experienced feelings of math anxiety. If you take a random sample of 10 students from this population, the mean and standard deviation of the number of students in the sample who have experienced math anxiety is:

• μ = 1.6; σ = 1.414

• μ = 1.6; σ = 1.265

• μ = 2; σ = 1.6

• μ = 2; σ = 1.414

• μ = 2; σ = 1.265

Which of the following is a true statement?

• The binomial setting requires that there are only two possible outcomes for each trial, while the geometric setting permits more than two outcomes.

• A geometric random variable takes on integer values from 0 to n.

• If X is a geometric random variable and the probability of success is 0.85, then the probability distribution of X will be skewed left, since 0.85 is closer to 1 than to 0.

• An important difference between binomial and geometric random variables is that there is a fixed number of trials in a binomial setting, and the number of trials varies in a geometric setting.

• The distribution of every binomial random variable is skewed right.

Jen’s commute to work requires that she take the Blue subway line, then transfer to the Red line. The length of the trip on the Blue line has a mean of 18 minutes with a standard deviation of 2 minutes. The Red line trip takes 12 minutes with a standard deviation of 1 minute. The waiting time between when she gets off the Blue line and her Red line train arrives has mean of 10 minutes and a standard deviation of 5 minutes. Assume (perhaps unrealistically) that these times are independent random variables. What are the mean and standard deviation of her entire commute?

• Mean = 40 minutes; Standard deviation = 8 minutes

• Mean = 40 minutes; Standard deviation = 5.48 minutes

• Mean = 40 minutes; Standard deviation = 2.83 minutes

• Mean = 30 minutes; Standard deviation = 5.48 minutes

• Mean = 30 minutes; Standard deviation = 8 minutes