AP Statistics Random Variables, Expected Value and Combining Variables Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Random Variables, Expected Value and Combining Variables – Maths PDF

Cover of the book AP Statistics Math Worksheets with Answers (PDF)

Download as a PDFThe AP Statistics worksheets book with answersPrintable PDFSee the book ›

Sommaire

This random variables answers for AP Statistics covers random variables, expected value and combining variables (Common Core AP STAT 4.7, AP STAT 4.8, AP STAT 4.9, HSS-MD.A.1, HSS-MD.A.2, HSS-MD.B.5), with a free PDF to print.

Each solution shows the products x times p, the squared deviations, and the combination rule used. All standard deviations of combinations assume independence, which is stated in the problem.

AP Statistics · Common Core State Standards: AP STAT 4.7, AP STAT 4.8, AP STAT 4.9, HSS-MD.A.1, HSS-MD.A.2, HSS-MD.B.5

Random variables answers: exercises 1 to 10

Exercise 1 – Valid distribution?

The probabilities must be between 0 and 1 and add to 1.

(a) Sum 1.1, no. (b) Sum 1.0, yes. (c) Contains -0.1, no.

Only (b) is valid.

Exercise 2 – Missing probability

$P(2)=1-0.1-0.4-0.2=0.3$.

0.3.

Exercise 3 – Find a probability

$P(X\ge 3)=0.20+0.15+0.10=0.45$. $P(X < 3)=0.30+0.25=0.55$.

0.45 and 0.55.

Exercise 4 – Expected value

Mean 2.50; variance 1.7500; standard deviation 1.323.

$1(0.30)+2(0.25)+3(0.20)+4(0.15)+5(0.10)=0.30+0.50+0.60+0.60+0.50=2.5$.

2.5 items.

Exercise 5 – Interpret the mean

The expected value is a long-run average.

Over many plays, the player loses an average of 50 cents per play.

Exercise 6 – Standard deviation

Mean 2.50; variance 1.7500; standard deviation 1.323.

Variance: $1.7500$. $\sigma_X=\sqrt{1.7500}=1.323$.

1.323 items: orders typically differ from the mean of 2.5 items by about 1.4 items.

Exercise 7 – Expected winnings

$4(0.2)+(-1)(0.8)=0.8-0.8=0$.

$0, so the game is fair.

Exercise 8 – Raffle

Net win: $500-5=495$ with probability 0.001; net loss $5 with probability 0.999.

$495(0.001)+(-5)(0.999)=0.495-4.995=-4.50$.

-$4.50 per ticket.

Exercise 9 – Linear transformation

$\mu_Y=3+2(2.5)=8$. $\sigma_Y=2(1.323)=2.646$.

Mean $8; standard deviation $2.65.

Exercise 10 – Sum of means

Mean: $40+25=65$. $\sigma=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5$.

Mean 65 minutes; standard deviation 5 minutes.

Exercises 11 and up

Exercise 11 – Difference

Mean: $40-25=15$. Standard deviation: $\sqrt{3^{2}+4^{2}}=5$.

Mean 15 minutes; standard deviation 5 minutes (same as the sum).

Exercise 12 – Standard deviations do not add

For independent variables, variances add, not standard deviations.

$\sigma=\sqrt{9+16}=5$, not 7.

Exercise 13 – 2X versus X + X

2X: $2(6)=12$. X + Y: $\sqrt{6^{2}+6^{2}}=\sqrt{72}\approx 8.49$.

12 compared with about 8.49: doubling one copy varies more than adding two independent copies.

Exercise 14 – Insurance

Profit for the insurer.

Profit: $400$, $400-3000=-2600$, $400-20000=-19600$.

$400(0.95)+(-2600)(0.04)+(-19600)(0.01)=380-104-196=80$.

$80 per customer per year.

Exercise 15 – Total of two games

$2+1=3$. $\sqrt{9+16}=5$.

Mean $3; standard deviation $5.

Exercise 16 – Weights

Mean: $10\times 12=120$. Variance: $10\times 0.25=2.5$. $\sigma=\sqrt{2.5}\approx 1.58$.

Mean 120 oz; standard deviation about 1.58 oz.

Exercise 17 – Normal approximation

$z=\frac{10-8}{1.414}=1.41$. $P(Z > 1.41)=0.0793$.

About 0.0793.

Exercise 18 – Change of units

$1.8(20)+32=68$. $1.8(4)=7.2$.

Mean 68°F; standard deviation 7.2°F.

Exercise 19 – Complete a distribution

Completed distribution.

$P(40)=1-0.9=0.1$. Mean: $10(0.4)+20(0.3)+30(0.2)+40(0.1)=4+6+6+4=20$.

P(40) = 0.1; mean 20.

Exercise 20 – Multi-step

$E(X)=0(0.5)+1(0.3)+2(0.2)=0.7$. Profit: $1.50X-0.40$, so $1.50(0.7)-0.40=0.65$.

$0.65 per minute.

Key facts behind this random variables answers

Key facts

Probabilities in a distribution are between 0 and 1 and sum to 1.
Mean (expected value): sum of x p; interpret it as the long-run average.
Standard deviation: square root of the sum of (x – mean)^2 p.
For a + bX: mean a + b mu, standard deviation |b| sigma.
For independent X, Y: means add or subtract; variances add (also for X – Y).

How to use this random variables answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the random variables answers as a PDF with the button below and keep it with your notes.

All AP Statistics chapters: AP Statistics math.

Voter.. post

Free printable PDF