1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years

*X* = the number of first-year students selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.

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 0.10 1,000,000 0.30 –1,000,000 0.60 Hardware Company *x**P*(*x*)3,000,000 0.20 1,000,000 0.40 –1,000,00 0.40 Biotech Firm *x**P*(*x*)6,00,000 0.10 0 0.70 –1,000,000 0.20 - $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 | 0.969 |

5 | $\frac{\text{250}}{\text{10,000}}$ = 0.025 |

25 | $\frac{\text{50}}{\text{10,000}}$ = 0.005 |

100 | $\frac{\text{10}}{\text{10,000}}$ = 0.001 |

Calculate the expected value of *X*.

0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35

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% of 32 questions correct. 75% 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% 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
- 0.0135
*P*(*x*= 12) = 0.3186*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*(25,0.40)- 10
- 0.0442
- 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 −0.0787. You lose about eight cents, on average, per game. - The house has the advantage.

*X*~*B*(15, 0.281)-
- Mean =
*μ*=*np*= 15(0.281) = 4.215 - Standard Deviation =
*σ*= $\sqrt{npq}$ = $\sqrt{15(0.281)(0.719)}$ = 1.7409

- Mean =
*P*(*x*> 5) = 1 –*P*(*x*≤ 5) = 1 – binomcdf(15, 0.281, 5) = 1 – 0.7754 = 0.2246*P*(*x*= 3) = binompdf(15, 0.281, 3) = 0.1927*P*(*x*= 4) = binompdf(15, 0.281, 4) = 0.2259

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

*X*= the number of adults in America who are surveyed until one says they will watch the Super Bowl.*X*~*G*(0.40)- 2.5
- 0.0187
- 0.2304

*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
- 0.9997
*X*= the number of pages we must survey until we find one that advertises footwear.*X*~*G*($\frac{29}{192}$)- 0.3881
- 6.6207 pages

*X*~*G*(0.25)-
- Mean =
*μ*= $\frac{1}{p}$ = $\frac{1}{0.25}$ = 4 - Standard Deviation = σ = $\sqrt{\frac{1-p}{{p}^{2}}}$ = $\sqrt{\frac{1-\text{0}\text{.25}}{{0.25}^{2}}}$ ≈ 3.4641

- Mean =
*P*(*x*= 10) = geometpdf(0.25, 10) = 0.0188*P*(*x*= 20) = geometpdf(0.25, 20) = 0.0011*P*(*x*≤ 5) = geometcdf(0.25, 5) = 0.7627

*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) ≈ 0.6860- There is a 15.7% 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 – 0.8944 = 0.1056

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

Using the Poisson distribution:

*μ*=*np*= 100(0.03) = 3*X*~*P*(3)*P*(*x*≤ 4) = poissoncdf(3, 4) ≈ 0.8153

Using the binomial distribution:

*X*~*B*(100, 0.03)*P*(*x*≤ 4) = binomcdf(100, 0.03, 4) ≈ 0.8179

The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

*X*= the number of fortune cookies that have an extra fortune- 0, 1, 2, 3,... 144
*X*~*B*(144, 0.03) or*P*(4.32)- 4.32
- 0.0124 or 0.0133
- 0.6300 or 0.6264
- As
*n*gets larger, the probabilities get closer together.