Below is our free AP Statistics unit 4 practice test. These questions focus on some very important background for statistics — probability. Simulation appears as a way to envision what might occur in a real-world situation if a certain assumption were true. The ideas of events and sets and union and intersection of sets are discussed , as well as expected value and standard deviation of random variables and functions of random variables.
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Question 1 |
A teacher rolls a die many times. The die is fair but the teacher doesn’t know it. The teacher wants to see how often a six is rolled. Using as many trials as are possible in the random number table below, what is the estimated probability of rolling a six?
3 8 5 4 8 3 2 2 0 1 9 5 8 6 2 5 5 5 6 1 9 3 5 0 5 3 1 5 9 7 4 0 9 2 2 8 4 9 2 2
0.05 | |
0.12 | |
0.17 | |
0.25 |
Question 1 Explanation:
The correct answer is (B). Any number 1–6 is a legitimate roll and is recorded. 0, 7, 8, 9 are not recorded. When we take out all the undesirable numbers we get:
$\dfrac{2}{27} = .12$
3 5 4 3 2 2 1 5 6 2 5 5 5 6 1 3 5 5 3 1 5 4 2 2 4 2 2
There are 27 rolls, and six is rolled twice:$\dfrac{2}{27} = .12$
Question 2 |
A student enters a drawing at a basketball game. There are 6 jerseys, 2 basketballs, and 20 free ticket packs. If 250 people enter the drawing, what is the probability the student will not win any prize?
1 | |
0.112 | |
0.08 | |
0.888 |
Question 2 Explanation:
The correct answer is (D). First, calculate the chance of winning:
$(6+2+20)÷250 = 0.112$
The student has a 0.112 chance of winning. Therefore the probability of not winning is:
$1 − 0.112 = 0.888$
$(6+2+20)÷250 = 0.112$
The student has a 0.112 chance of winning. Therefore the probability of not winning is:
$1 − 0.112 = 0.888$
Question 3 |
Which of the following are mutually exclusive events?
Being from Florida and knowing how to ice skate, because no one from Florida knows how to ice skate. | |
Rolling an odd number on a die and rolling a number less than three, because there is no overlap. | |
A flipped coin landing on heads and a flipped coin landing on tails, because these events cannot both occur on the same trial. | |
Drawing a heart from a deck of cards and drawing a face card, because these are independent events. |
Question 3 Explanation:
The correct answer is (C). These two events cannot occur simultaneously; therefore, they are mutually exclusive.
Question 4 |
In a certain middle school, 28% of students play soccer and basketball, and 40% of students play basketball. What is the probability that a student plays soccer, given that the student plays basketball?
0.112 | |
0.12 | |
0.28 | |
0.7 |
Question 4 Explanation:
The correct answer is (D). The formula for conditional probability is:
$P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)}$
$P(A)$ is the probability of the first event and $P(A \text{ and } B)$ is the probability of both events happening at the same time (playing soccer and playing basketball).
$P(B|A)$ is the probability of event $B$ when it is known that event $A$ is the case.
$\dfrac{0.28}{0.40} = 0.7$
$P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)}$
$P(A)$ is the probability of the first event and $P(A \text{ and } B)$ is the probability of both events happening at the same time (playing soccer and playing basketball).
$P(B|A)$ is the probability of event $B$ when it is known that event $A$ is the case.
$\dfrac{0.28}{0.40} = 0.7$
Question 5 |
A student surveys students in a school to determine the rate of food allergies. The student finds that, out of a sample of 200 students, 5 have a peanut allergy, 2 have a tree nut allergy, 1 has a shellfish allergy, and 4 have an egg allergy.
Assuming each of these events are independent of the others is the probability that a student has either a peanut or tree nut allergy?
0.00025 | |
0.01 | |
0.025 | |
0.03475 |
Question 5 Explanation:
The correct answer is (D). The probability of the union of events (or) is calculated by adding the probability of each event and subtracting the probability of both events occurring at once. In this case, that is the product of the probabilities because the events are independent:
$\dfrac{5}{200} + \dfrac{2}{200} − \left(\dfrac{5}{200} \ast \dfrac{2}{200}\right)$
$0.025+0.01-(0.025*0.01)$
$= 0.03475$
$\dfrac{5}{200} + \dfrac{2}{200} − \left(\dfrac{5}{200} \ast \dfrac{2}{200}\right)$
$0.025+0.01-(0.025*0.01)$
$= 0.03475$
Question 6 |
A researcher surveyed 200 adults. Based on the results, the probability of a person taking a given number of pizza slices is displayed in the table below.
Number of slices |
Probability |
1 |
0.450 |
2 |
0.405 |
3 |
0.130 |
4 |
0.010 |
5 |
0.005 |
Based on this data, what is the average number of slices taken by a single individual?
1.715 | |
1.805 | |
1.940 | |
2.015 |
Question 6 Explanation:
The correct answer is (A). The average number of slices will be the sum of all possible outcomes multiplied by their probabilities:
1 × .450 + 2 × .405 + 3 × .130 + 4 × .010 + 5 × .005 = 1.715
1 × .450 + 2 × .405 + 3 × .130 + 4 × .010 + 5 × .005 = 1.715
Question 7 |
At a craft fair, the average purchase amount is \$30 with a standard deviation of \$2. In addition, there is 7% sales tax plus a \$5 charge to get in to the fair. Which of the following statements are true?
The mean and standard deviation of purchases with tax and entry are \$37.10 and \$2.14 | |
The mean and standard deviation of purchases with tax and entry are \$35 and \$2. | |
The mean and standard deviation of purchases with tax and entry are \$37.10 and \$2. | |
The mean and standard deviation of purchases with tax and entry are \$32.10 and \$0.14. |
Question 7 Explanation:
The correct answer is (A). The transformed equation for this situation is:
$1.07X+5$
Therefore the mean is:
$1.07*30+5=\$37.10$
The standard deviation is:
$1.07*2 = \$2.14$
$1.07X+5$
Therefore the mean is:
$1.07*30+5=\$37.10$
The standard deviation is:
$1.07*2 = \$2.14$
Question 8 |
A student rolls a die 36 times, with a success defined as rolling a 5 or 6. What are the mean and standard deviation for the distribution of the number of successes?
Mean = 3.5, Standard Deviation = 1 | |
Mean = 12, Standard Deviation ≈2.828 | |
Mean = 3.5, Standard Deviation ≈ 2.828 | |
Mean = 12, Standard Deviation = 1 |
Question 8 Explanation:
The correct answer is (B). For a binomial distribution, the mean is $n \ast p$, where $n$ = number of trials and $p$ = probability of success:
$36 * \dfrac{1}{3} = 12$
The standard deviation is:
$\sqrt{np(1-p).}$ $\sqrt{36 * \dfrac{1}{3} (1 - \dfrac{1}{3})} = \sqrt{8} ≈ 2.828$
$36 * \dfrac{1}{3} = 12$
The standard deviation is:
$\sqrt{np(1-p).}$ $\sqrt{36 * \dfrac{1}{3} (1 - \dfrac{1}{3})} = \sqrt{8} ≈ 2.828$
Question 9 |
An automated dialer is dialing random numbers until it reaches a valid number. The mean for this situation is 12 and the standard deviation is approximately 11.5. Which statement is true?
The expected number of attempts before a valid number is reached is 11.5. | |
There will be approximately 0.5 valid numbers for every 12 attempts. | |
The expected number of attempts before a valid number is reached is 12. | |
The expected number of attempts before two valid numbers are reached is 23.5. |
Question 9 Explanation:
The correct answer is (C). In a geometric distribution, the mean is the expected number of attempts before a success.
Question 10 |
An online survey found that 27 percent of students age 18 and over at a local community college are the first in their family to attend college. A random sample of 40 students from the community college who are age 18 or over is selected. Let the random variable B represent the number of students in the sample who are the first in their family to attend college.
What is the probability that B will equal 30?
$\begin{pmatrix}
40 \\
30 \\
\end{pmatrix}
(0.27)^{30}(0.73)^{10}$ | |
$\begin{pmatrix}
40 \\
30 \\
\end{pmatrix}
(0.27)^{10}(0.73)^{30}$ | |
$\begin{pmatrix}
40 \\
10 \\
\end{pmatrix}
(0.27)^{10}(0.73)^{30}$ | |
$\begin{pmatrix}
40 \\
27 \\
\end{pmatrix}
(0.75)^{30}(0.25)^{10}$ | |
$30(0.27)^{40}$ |
Question 10 Explanation:
The correct answer is (A). Success is defined as selecting a person who is the first in their family to attend college and failure is defined as selecting a person who is not the first in their family to attend college. Random variable B has a binomial distribution, with probability of success of 0.27 and probability of failure of (1 − 0.27) = 0.73. Using the binomial formula with $n$ = 40, $x$ = 30, $p$ = 0.27, $q$ = 0.73
$P(x) = \begin{pmatrix} n \\ x \\ \end{pmatrix} p^{x} q^{n-x}$
The probability of 30 out of 40 people being the first in their family to attend college is given by
$\begin{pmatrix} 40 \\ 30 \\ \end{pmatrix} (0.27)^{30}(0.73)^{10}$
$P(x) = \begin{pmatrix} n \\ x \\ \end{pmatrix} p^{x} q^{n-x}$
The probability of 30 out of 40 people being the first in their family to attend college is given by
$\begin{pmatrix} 40 \\ 30 \\ \end{pmatrix} (0.27)^{30}(0.73)^{10}$
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