Question

remember, each event is equally likely to occur (probability is 1/16 or 0.0625). fill in the following table. (look at table 5 - 2 on page 209 if you need help)
what is the sum of the numbers in the ways this can occur column? does this make sense?
what is the sum of the numbers in the p(x) column? does this make sense?
fill in the following table. (look at table 5 - 3 on page 212 if you need help)
what is the mean or expected value?
what is the variance?
what is the standard deviation?
make a histogram by hand or using excel as you did in a previous project plotting x (horizontal axis) and probability (vertical axis).

Explanation:

Step1: Calculate the number of ways for each \(x\)

• For \(x = 0\): \(C(4,0)=\frac{4!}{0!(4 - 0)!}=1\)
• For \(x = 1\): \(C(4,1)=\frac{4!}{1!(4 - 1)!}=4\)
• For \(x = 2\): \(C(4,2)=\frac{4!}{2!(4 - 2)!}=6\)
• For \(x = 3\): \(C(4,3)=\frac{4!}{3!(4 - 3)!}=4\)
• For \(x = 4\): \(C(4,4)=\frac{4!}{4!(4 - 4)!}=1\)

Step2: Calculate \(P(x)\) for each \(x\)

Since the total number of outcomes when tossing 4 coins is \(2^4 = 16\), \(P(x)=\frac{C(4,x)}{16}\)

• For \(x = 1\): \(P(1)=\frac{4}{16}=0.25\)
• For \(x = 2\): \(P(2)=\frac{6}{16}=0.375\)
• For \(x = 3\): \(P(3)=\frac{4}{16}=0.25\)

Step3: Calculate \(x\cdot P(x)\)

• For \(x = 1\): \(1\times0.25 = 0.25\)
• For \(x = 2\): \(2\times0.375=0.75\)
• For \(x = 3\): \(3\times0.25 = 0.75\)

Step4: Calculate \(\mu\) (mean)

\(\mu=\sum x\cdot P(x)=0\times0.0625 + 1\times0.25+2\times0.375 + 3\times0.25+4\times0.0625\)
\(=0 + 0.25+0.75 + 0.75+0.25=2\)

Step5: Calculate \((x-\mu)^2\cdot P(x)\)

• For \(x = 0\): \((0 - 2)^2\times0.0625=4\times0.0625 = 0.25\)
• For \(x = 1\): \((1 - 2)^2\times0.25=1\times0.25 = 0.25\)
• For \(x = 2\): \((2 - 2)^2\times0.375=0\times0.375 = 0\)
• For \(x = 3\): \((3 - 2)^2\times0.25=1\times0.25 = 0.25\)
• For \(x = 4\): \((4 - 2)^2\times0.0625=4\times0.0625 = 0.25\)

Step6: Calculate variance \(\sigma^2\)

\(\sigma^2=\sum(x-\mu)^2\cdot P(x)=0.25+0.25 + 0+0.25+0.25 = 1\)

Step7: Calculate standard deviation \(\sigma\)

\(\sigma=\sqrt{\sigma^2}=\sqrt{1}=1\)

Step8: Sum of “ways” column

\(1 + 4+6 + 4+1=16\). This makes sense as \(2^4 = 16\) (total number of outcomes when tossing 4 coins)

Step9: Sum of \(P(x)\) column

\(0.0625+0.25 + 0.375+0.25+0.0625 = 1\). This makes sense as the sum of all probabilities in a probability distribution is 1

Answer:

• Sum of “ways” column: \(16\)
• Sum of \(P(x)\) column: \(1\)
• Mean (\(\mu\)): \(2\)
• Variance (\(\sigma^2\)): \(1\)
• Standard deviation (\(\sigma\)): \(1\)