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AP Statistics : Statistical Patterns and Random Phenomena

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Example Questions

Example Question #71 : Statistical Patterns And Random Phenomena

A basketball player makes $40\%$ of his three-point shots. If he takes three-point shots each game, how many points per game does he score from three-point range?

Possible Answers:

Correct answer:

Explanation:

First convert $40\% \rightarrow 0.40$ .

The player's three-point shooting follows a binomial distribution with n = 10 and p = 0.4 .

On average, he thus makes $np = 10\cdot 0.4 = 4$ three-point shots per game.

This means he averages 12 points per game from three-point range if he tries to make 10 three-pointers per game.

4(3)=12

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Example Question #11 : Random Variables

Tim samples the average plant height of potato plants for his science class and finds the following distribution (in inches):

4,4,5,5,6,6,6,6,6,7,7,7,8,9,10,10,12,14

Which of the following is/are true about the data?

i: the mode is

ii: the mean is 7.33333

iii: the median is

iv: the range is

Possible Answers:

all of the above

i & iii

i, ii & iii

i & ii

ii, iii & iv

Correct answer:

i & ii

Explanation:

Analyzing the data, there are more 6s than anything else (mode), the median is between and (6.5) , the mean is 132/18=7.33333 , and the range is 14-4=10

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Example Question #71 : Probability

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the mean outcome for the number of hours that Robert will work?

Possible Answers:

31.6

30.5

122

0.25

32.7

Correct answer:

32.7

Explanation:

We are required to find the mean outcome where the probability of each possible result varies--the random/weighted mean. First, multiply each possible outcome by the probability of that outcome occurring. Second, add these results together.

$(45\cdot 0.3) + (40\cdot 0.2) + (25\cdot 0.4) + (12\cdot 0.1) = 32.7$

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Example Question #1 : How To Find Standard Deviation Of A Random Variable

We have two independent, normally distributed random variables $\small X_1$ and $\small X_2$ such that $\small X_1$ has mean $\small 0$ and variance $\small 2$ and $\small X_2$ has mean $\small 0$ and variance $\small 1$ . What is the probability distribution of the difference of the random variables, $\small \small X_1-X_2$ ?

Possible Answers:

Normal distribution with mean $\small 0$ and variance $\small 3$ .

Normal distribution with mean $\small 0$ and variance $\small 1$ .

Normal distribution with mean $\small 1$ and variance $\small 2$ .

Normal distribution with mean $\small 1$ and variance $\small 3$ .

Correct answer:

Normal distribution with mean $\small 0$ and variance $\small 3$ .

Explanation:

The mean for any set of random variables is additive in the sense that

$\small Mean(X_1+X_2)=Mean(X_1)+Mean(X_2)$

The difference is also additive, so we have

$\small Mean(X_1-X_2)=Mean(X_1)-Mean(X_2)$

This means the mean of $\small X_1-X_2$ is $\small 0$ .

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers $\small a,b$ (even when negative), we have

$\small Var(aX_1+bX_2)=a^2Var(X_1)+b^2Var(X_2)$ .

So for this difference, we have

$\small \small \small Var(X_1-X_2)=Var(X_1+(-1)X_2)=1^2Var(X_1)+(-1)^2Var(X_2)$

$\small =Var(X_1)+Var(X_2)=2+1=3$ .

So the mean and variance are $\small 0$ and $\small 3$ , respectively. In addition to that, $\small X_1-X_2$ is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.

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Example Question #2 : How To Find Standard Deviation Of A Random Variable

If $\small X_1$ and $\small X_2$ are two independent random variables with Var(X_1)=3 and Var(X_2)=3 , what is the standard deviation of the sum, $\small X_1+X_2?$

Possible Answers:

$\small 6$

${}\small \small \sqrt{3}$

$\small \sqrt{6}$

$\small \small \sqrt{18}$

Correct answer:

$\small \sqrt{6}$

Explanation:

If the random variables are independent, the variances are additive in the sense that

$\small Var(X+Y)=Var(X)+Var(Y)$ .

So then the variance of the sum is

$\small \small Var(X_1+X_2)=Var(X_1)+Var(X_2)=3+3=6$ .

The standard deviation is the square root of the variance, so we have

$\small SD(X_1+X_2)=\sqrt{6}$ .

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Example Question #1 : How To Find Standard Deviation Of A Random Variable

Consider the discrete random variable $\small X$ that takes the following values with the corresponding probabilities:

$\small X=1$ with $\small P(X=1)=\frac{1}{8}$
$\small X=2$ with $\small P(X=2)=\frac{3}{8}$
$\small X=3$ with $\small P(X=3)=\frac{1}{2}$

Compute the probability $\small P(X\ge2)$ .

Possible Answers:

$\small \frac{7}{8}$

$\small \frac{1}{2}$

$\small \frac{1}{4}$

$\small \frac{1}{8}$

Correct answer:

$\small \frac{7}{8}$

Explanation:

This probability is simple to compute:

We want to add the probability that X is greater or equal to two. This means the probability that X=2 or X=3.

$\small \small P(X\ge 2)=P(X=2 \or X=3)$

Adding the necessary probabilities we arrive at the solution.

=P(X=2)+P(X=3)

$\small =\frac{3}{8}+\frac{1}{2}=\frac{7}{8}$

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Example Question #2 : How To Find Standard Deviation Of A Random Variable

Consider the discrete random variable $\small X$ that takes the following values with the corresponding probabilities:

$\small X=1$ with $\small \small P(X=1)=\frac{1}{4}$
$\small X=2$ with $\small \small P(X=2)=\frac{1}{4}$
$\small X=3$ with $\small \small P(X=3)=\frac{1}{4}$
$\small X=4$ with $\small P(X=4)=\frac{1}{4}$

Compute the expected value of the distribution.

Possible Answers:

$\small 2$

$\small 3$

$\small 10$

$\small 2.5$

Correct answer:

$\small 2.5$

Explanation:

The expected value is computed as

$\small \mathbb{E}[X]=\small \sum_{x} xP(X=x)$

for any values of x that the random variable takes.

So we have

$\small \small \mathbb{E}[X]= \frac{1}{4}\cdot 1+\frac{1}{4}\cdot 2+\frac{1}{4}\cdot 3+\frac{1}{4}\cdot 4$

$\small =\frac{10}{4}=2.5$

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Example Question #4 : How To Find Standard Deviation Of A Random Variable

The average number of calories in a Lick Yo' Lips lollipop is 210 , with a standard deviation of . The calories per lollipop are normally distributed, so what percent of lollipops have more than 225 calories?

Possible Answers:

$6.68\%$

$47.9\%$

$3.56\%$

$4.57\%$

$45.7\%$

Correct answer:

$6.68\%$

Explanation:

The random variable X = number of calories per lollipop, so the answer is

$P(\mbox{Number of Calories} > 225)$ or

$P(X > 225) = P(Z > \frac{(225-210)}{10}) = P(Z > 1.5) = 1- 0.9332 = 0.0668.$

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Example Question #1 : How To Find Standard Deviation Of A Random Variable

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the standard deviation of the possible outcomes?

Possible Answers:

32.7

11.07

13.5

35.8

122.61

Correct answer:

11.07

Explanation:

There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together. Fourth, find the square root of the result.

$(45-32.7)^{2}(0.3)=45.38$

$(40-32.7)^{2}(0.2)=10.66$

$(25-32.7)^{2}(0.4)=23.72$

$(12-32.7)^{2}(0.1)=42.85$

45.38+10.66+23.72+42.85=122.61

$\sqrt{}122.61=11.07$

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Example Question #72 : Statistical Patterns And Random Phenomena

Possible Answers:

Correct answer:

Explanation:

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AP Statistics : Statistical Patterns and Random Phenomena

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #71 : Statistical Patterns And Random Phenomena

Example Question #11 : Random Variables

Example Question #71 : Probability

Example Question #1 : How To Find Standard Deviation Of A Random Variable

Example Question #2 : How To Find Standard Deviation Of A Random Variable

Example Question #1 : How To Find Standard Deviation Of A Random Variable

Example Question #2 : How To Find Standard Deviation Of A Random Variable

Example Question #4 : How To Find Standard Deviation Of A Random Variable

Example Question #1 : How To Find Standard Deviation Of A Random Variable

Example Question #72 : Statistical Patterns And Random Phenomena

All AP Statistics Resources

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