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4.12: Chapter Solution (Practice + Homework)

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    1.

    \(x\) \(P(x)\)
    0 0.12
    1 0.18
    2 0.30
    3 0.15
    4 0.10
    5 0.10
    6 0.05
    Table \(\PageIndex{1}\)

    3. 0.10 + 0.05 = 0.15

    5. 1

    7. 0.35 + 0.40 + 0.10 = 0.85

    9. 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45

    11.

    \(x\) \(P(x)\)
    0 0.03
    1 0.04
    2 0.08
    3 0.85
    Table \(\PageIndex{2}\)

    13. Let \(X =\) the number of events Javier volunteers for each month.

    15.

    \(x\) \(P(x)\)
    0 0.05
    1 0.05
    2 0.10
    3 0.20
    4 0.25
    5 0.35
    Table \(\PageIndex{3}\)

    17. 1 – 0.05 = 0.95

    18. \(X =\) the number of business majors in the sample.

    19. 2, 3, 4, 5, 6, 7, 8, 9

    20. \(X =\) the number that reply “yes”

    22. 0, 1, 2, 3, 4, 5, 6, 7, 8

    24. 5.7

    26. 0.4151

    28. \(X =\) the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.

    30. 1,2,…

    32. 1.4

    35. 0, 1, 2, 3, 4, …

    37. 0.0485

    39. 0.0214

    41. \(X =\) the number of U.S. teens who die from motor vehicle injuries per day.

    43. 0, 1, 2, 3, 4, ...

    45. No

    48.

    1. X = the number of pages that advertise footwear
    2. 0, 1, 2, 3, ..., 20
    3. 3.03
    4. 1.5197

    50.

    1. X = the number of Patriots picked
    2. 0, 1, 2, 3, 4
    3. Without replacement

    53. \(X =\) the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, ...25

    55. 0.0165

    57.

    1. X = the number of DVDs a Video to Go customer rents
    2. 0.12
    3. 0.11
    4. 0.77

    59. 4.43

    61. c

    63.

    • X = number of questions answered correctly
    • X ~ B(32, 13)(32, 13)
    • 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."
    • 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.

    65.

    1. X = the number of college and universities that offer online offerings.
    2. 0, 1, 2, …, 13
    3. X ~ B(13, 0.96)
    4. 12.48
    5. 0.0135
    6. P(x = 12) = 0.3186 P(x = 13) = 0.5882 More likely to get 13.

    67.

    • 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.

    69.

    1. X = the number of audits in a 20-year period
    2. 0, 1, 2, …, 20
    3. X ~ B(20, 0.02)
    4. 0.4
    5. 0.6676
    6. 0.0071

    71.

    1. X = the number of matches
    2. 0, 1, 2, 3
    3. In dollars: −1, 1, 2, 3
    4. 1212
    5. The answer is −0.0787. You lose about eight cents, on average, per game.
    6. The house has the advantage.

    73.

    1. X ~ B(15, 0.281)
      This histogram shows a binomial probability distribution. It is made up of bars that are fairly normally distributed. The x-axis shows values from 0 to 15, with bars from 0 to 9. The y-axis shows values from 0 to 0.25 in increments of 0.05.
      1. Mean = μ = np = 15(0.281) = 4.215
      2. Standard Deviation = σ = npq−−−√𝑛𝑝𝑞 = 15(0.281)(0.719)−−−−−−−−−−−−−√15(0.281)(0.719) = 1.7409
    2. P(x > 5)=1 – 0.7754 = 0.2246
      P(x = 3) = 0.1927
      P(x = 4) = 0.2259
      It is more likely that four people are literate that three people are.

    75.

    1. X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
    2. X ~ G(0.40)
    3. 2.5
    4. 0.0187
    5. 0.2304

    77.

    1. X = the number of pages that advertise footwear
    2. X takes on the values 0, 1, 2, ..., 20
    3. X ~ B(20, 2919229192)
    4. 3.02
    5. No
    6. 0.9997
    7. X = the number of pages we must survey until we find one that advertises footwear. X ~ G(2919229192)
    8. 0.3881
    9. 6.6207 pages

    79. 0, 1, 2, and 3

    81.

    1. X ~ G(0.25)
      1. Mean = μ = 1p1𝑝 = 10.2510.25 = 4
      2. Standard Deviation = σ = 1−pp2−−−√1−𝑝𝑝2 = 1−0.250.252−−−−−√1−0.250.252 ≈ 3.4641
    2. P(x = 10) = 0.0188
    3. P(x = 20) = 0.0011
    4. P(x ≤ 5) = 0.7627

    82.

    1. X ~ P(5.5); μ = 5.5; σ = 5.5−−−√𝜎 = 5.5 ≈ 2.3452
    2. P(x ≤ 6) ≈ 0.6860
    3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
    4. P(x > 8) = 1 – P(x ≤ 8) ≈ 1 – 0.8944 = 0.1056

    84.

    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) ≈ 0.8153

    Using the binomial distribution:

    • X ~ B(100, 0.03)
    • P(x ≤ 4) = 0.8179

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

    86.

    1. X = the number of children for a Spanish woman
    2. 0, 1, 2, 3,...
    3. 0.2299
    4. 0.5679
    5. 0.4321

    88.

    1. X = the number of fortune cookies that have an extra fortune
    2. 0, 1, 2, 3,... 144
    3. 4.32
    4. 0.0124 or 0.0133
    5. 0.6300 or 0.6264
    6. As n gets larger, the probabilities get closer together.

    90.

    1. X = the number of people audited in one year
    2. 0, 1, 2, ..., 100
    3. 2
    4. 0.1353
    5. 0.3233

    92.

    1. X = the number of shell pieces in one cake
    2. 0, 1, 2, 3,...
    3. 1.5
    4. 0.2231
    5. 0.0001
    6. Yes

    94. d


    4.12: Chapter Solution (Practice + Homework) is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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