1(.35) + 2(.20) + 3(.15) + 4(.15) + 5(.10) + 6(.05) = .35 + .40 + .45 + .60 + .50 + .30 = 2.6 years

*X* = the number of freshmen selected from the study until one replied *yes* to the law that was passed.

The variable of interest is *X*, or the gain or loss, in dollars.

The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards.

We first need to construct the probability distribution for *X*. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of *X* to determine the expected value.

Card Event | X net gain/loss |
P(X) |
---|---|---|

Face Card and Heads | 6 | $\left(\frac{12}{52}\right)\left(\frac{1}{2}\right)=\left(\frac{6}{52}\right)$ |

Face Card and Tails | 2 | $\left(\frac{12}{52}\right)\left(\frac{1}{2}\right)=\left(\frac{6}{52}\right)$ |

(Not Face Card) and (H or T) | –2 | $\left(\frac{40}{52}\right)\left(1\right)=\left(\frac{40}{52}\right)$ |

- $\text{Expectedvalue}=(6)\left(\frac{6}{52}\right)+(2)\left(\frac{6}{52}\right)+(-2)\left(\frac{40}{52}\right)=\u2013\frac{32}{52}$
- Expected value = –$0.62, rounded to the nearest cent
- If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average.
- You should not play this game to win money because the expected value indicates an expected average loss.

Software Company *x**P*(*x*)5,000,000 .10 1,000,000 .30 –1,000,000 .60 Hardware Company *x**P*(*x*)3,000,000 .20 1,000,000 .40 –1,000,00 .40 Biotech Firm *x**P*(*x*)6,000,000 .10 0 .70 –1,000,000 .20 - $200,000; $600,000; $400,000
- third investment because it has the lowest probability of loss
- first investment because it has the highest probability of loss
- second investment

Let *X* = the amount of money to be won on a ticket. The following table shows the PDF for *X*:

x |
P(x) |
---|---|

0 | .969 |

5 | $\frac{\text{250}}{\text{10,000}}$ = .025 |

25 | $\frac{\text{50}}{\text{10,000}}$ = .005 |

100 | $\frac{\text{10}}{\text{10,000}}$ = .001 |

Calculate the expected value of *X*.

0(.969) + 5(.025) + 25(.005) + 100(.001) = .35

A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.

*X* = the number of patients calling in claiming to have the flu, who actually have the flu.

*X* = 0, 1, 2, ...25

*X*= number of questions answered correctly*X*~*B*$\left(\text{32,}\frac{\text{1}}{\text{3}}\right)$- We are interested in MORE THAN 75 percent of 32 questions correct. 75 percent of 32 is 24. We want to find
*P*(*x*> 24). The event*more than 24*is the complement of*less than or equal to 24*. - Using your calculator's distribution menu: 1 – binomcdf$\left(\text{32,}\frac{\text{1}}{\text{3}},\text{24}\right)$
*P*(*x*> 24) = 0- The probability of getting more than 75 percent of the 32 questions correct when randomly guessing is very small and practically zero.

*X*= the number of college and universities that offer online offerings.- 0, 1, 2, …, 13
*X*~*B*(13, 0.96)- 12.48
- .0135
*P*(*x*= 12) = .3186*P*(*x*= 13) = 0.5882 More likely to get 13.

*X*= the number of fencers who do*not*use the foil as their main weapon- 0, 1, 2, 3,... 25
*X*~*B*(25,.40)- 10
- .0442
- The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

*X*= the number of matches- 0, 1, 2, 3
*X*~*B*$\left(3,\frac{1}{6}\right)$- In dollars: −1, 1, 2, 3
- \frac{1}{2}
- Multiply each
*Y*value by the corresponding*X*probability from the PDF table. The answer is −.0787. You lose about eight cents, on average, per game. - The house has the advantage.

*X*~*B*(15, .281)-
- Mean =
*μ*=*np*= 15(.281) = 4.215 - Standard Deviation =
*σ*= $\sqrt{npq}$ = $\sqrt{15(.281)(.719)}$ = 1.7409

- Mean =
*P*(*x*> 5) = 1 –*P*(*x*≤ 5) = 1 – binomcdf(15, .281, 5) = 1 – 0.7754 = .2246*P*(*x*= 3) = binompdf(15, .281, 3) = .1927*P*(*x*= 4) = binompdf(15, .281, 4) = .2259

It is more likely that four people are literate than three people are.

*X*= the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.*X*~*G*(.40)- 2.5
- .0187
- .2304

*X*= the number of pages that advertise footwear*X*takes on the values 0, 1, 2, ..., 20*X*~*B*(20, $\frac{29}{192}$)- 3.02
- no
- .9997
*X*= the number of pages we must survey until we find one that advertises footwear.*X*~*G*($\frac{29}{192}$)- .3881
- 6.6207 pages

*X*~*G*(.25)-
- mean =
*μ*= $\frac{1}{p}$ = $\frac{1}{0.25}$ = 4 - standard deviation = σ = $\sqrt{\frac{1-p}{{p}^{2}}}$ = $\sqrt{\frac{1-\text{.25}}{{.25}^{2}}}$ ≈ 3.4641

- mean =
*P*(*x*= 10) = geometpdf(.25, 10) = .0188*P*(*x*= 20) = geometpdf(.25, 20) = .0011*P*(*x*≤ 5) = geometcdf(.25, 5) = .7627

*X*= the number of pages that advertise footwear- 0, 1, 2, 3, ..., 20
*X*~*H*(29, 163, 20),*r*= 29,*b*= 163,*n*= 20- 3.03
- 1.5197

*X*~*P*(5.5);*μ*= 5.5; $\sigma \text{=}\sqrt{5.5}$ ≈ 2.3452*P*(*x*≤ 6) = poissoncdf(5.5, 6) ≈ .6860- There is a 15.7 percent probability that the law staff will receive more calls than they can handle.
*P*(*x*> 8) = 1 –*P*(*x*≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – .8944 = .1056

Let *X* = the number of defective bulbs in a string.

Using the Poisson distribution:

*μ*=*np*= 100(.03) = 3*X*~*P*(3)*P*(*x*≤ 4) = poissoncdf(3, 4) ≈ .8153

Using the binomial distribution

*X*~*B*(100, .03)*P*(*x*≤ 4) = binomcdf(100, .03, 4) ≈ .8179

The Poisson approximation is very good—the difference between the probabilities is only .0026.

*X*= the number of fortune cookies that have an extra fortune- 0, 1, 2, 3,... 144
*X*~*B*(144, .03) or*P*(4.32)- 4.32
- .0124 or .0133
- .6300 or .6264
- As
*n*gets larger, the probabilities get closer together.

- You can use randInt (0,1,5) to generate five trials of the experiment. Count the number of 1’s generated to determine the number of successes.
- Student answers may vary.
- Student answers may vary.
- The theoretical mean is $\left(5\right)\left(.5\right)=2.5$. The theoretical standard deviation is $\sqrt{\left(5\right)\left(.5\right)\left(0.5\right)}=\sqrt{1.25}$.