Search Header Logo

AP Stats Review U4-5

Authored by Mary Kandigian

Mathematics

12th Grade

CCSS covered

Used 24+ times

AP Stats Review U4-5
AI

AI Actions

Add similar questions

Adjust reading levels

Convert to real-world scenario

Translate activity

More...

    Content View

    Student View

8 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

A company that ships crystal bowls claims that bowls arrive undamaged in 95 percent of the shipments. Let the random variable G represent the number of shipments with undamaged bowls in 25 randomly selected shipments. Random variable G follows a binomial distribution with a mean of 23.75 shipments and a standard deviation of approximately 1.09 shipments. Which of the following is the best interpretation of the mean?

Every shipment of 25 bowls will have 23.75 undamaged bowls

Every shipment of 25 bowls will have 23.75 damaged bowls

On average, the company receives 23.75 shipments before receiving the first shipment with a damaged bowl.

For all possible shipments of 25, the average number of damaged shipments is equal to 23.75

For all possible shipments of size 25, the average number of undamaged shipments is equal to 23.75.

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Suppose the variance in trunk diameter of the giant sequoia tree species is 15.7 sq. meters, while the variance in trunk diameter of the California redwood tree species is 10.6 sq. meters. Let x-bar1 represent the average trunk diameter of 4 randomly sampled giant sequoia trees, and let x-bar2 represent the average trunk diameter of 3 randomly sampled California redwood trees. If the random sampling is done with replacement, what is the standard deviation of x-bar1 - x-bar2 for the sampling distribution of the difference in sample means x-bar1 - x-bar 2?

 15.7410.63\sqrt{\frac{15.7}{4}-\frac{10.6}{3}}  

 15.7  10.6\sqrt{15.7\ -\ 10.6}  

 15.74+10.63\sqrt{\frac{15.7}{4}+\frac{10.6}{3}}  

 15.74+10.63\sqrt{\frac{15.7}{4}}+\sqrt{\frac{10.6}{3}}  

 15.74 +10.63\frac{15.7}{4\ }+\frac{10.6}{3}  

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

There were 5.317 previously owned homes sold in a western city in the year 2000. The distribution of sales prices of these homes was strongly right-skewed, with a mean of $206,274 and a standard deviation of $37,881. If all possible simple random samples of size 100 are drawn from this population and the mean is computed for each of these samples, which of the following describes the sampling distribution of the sample mean?

Approximately normal with mean $206,274 and standard deviation $3,788

Approximately normal with mean $206,274 and standard deviation $37,881

Approximately normal with mean $206,274 and standard deviation $520

Strongly right-skewed with mean $206,274 and standard deviation $3,788

Strongly right-skewed with mean $206,274 and standard deviation $37,881

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Consider a data set of positive values, at least two of which are not equal. Which of the following sample statistics will be changed when each value of this data set is multiplied by a constant whose absolute value is greater than 1? I. The mean; II. The median; III The standard deviation

I only

II only

III only

I and II only

I, II, and III

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Joe and Matthew plan to visit a bookstore. Based on their previous visits to the bookstore, the probability distributions of the number of books they will buy are given below. For Joe P(X = 0) = 0.50; P(X = 1) = 0.25; P(X = 2) = 0.25. For Matthew P(X = 0) = 0.25; P(X = 1) = 0.50; P(X = 2) = 0.25. Assuming that Joe and Matthew make their decisions independently, what is the probability that they will purchase no books on this visit to the bookstore?

0.0625

0.1250

0.1875

0.2500

0.7500

6.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Sixty-five percent of all divorce cases cite incompatibility as the underlying reason. If four couples file for a divorce, what is the probability that exactly two will state incompatibility as the reason?

2(0.65)(0.35

2(0.65)2(0.35)22\left(0.65\right)^2\left(0.35\right)^2

4(0.65)2(0.35)24\left(0.65\right)^2\left(0.35\right)^2

6(0.65)2(035)26\left(0.65\right)^2\left(035\right)^2

0.650.65

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

It is estimated that two out of five high school students would fall victim to a phishing email (an online scam asking for sensitive information) if it appears to originate from their high school main office. What is the probability that the first student to fall victim will be the third student who is sent a phishing email that appears to originate from their high school main office?

(0.4)3\left(0.4\right)^3

(0.6)3\left(0.6\right)^3

(0.6)(0.4)2\left(0.6\right)\left(0.4\right)^2

(0.6)2(0.4)\left(0.6\right)^2\left(0.4\right)

10(0.6)2(0.4)10\left(0.6\right)^2\left(0.4\right)

Tags

CCSS.HSS.MD.B.6

CCSS.HSS.MD.B.7

Access all questions and much more by creating a free account

Create resources

Host any resource

Get auto-graded reports

Google

Continue with Google

Email

Continue with Email

Classlink

Continue with Classlink

Clever

Continue with Clever

or continue with

Microsoft

Microsoft

Apple

Apple

Others

Others

Already have an account?