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5.11: Chapter Solution (Practice + Homework)

  • Page ID
    5570
  • This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle.
    Figure \(\PageIndex{38}\)

    Figure \(\PageIndex{41}\)

  • This is a conditional probability question. \(P(x > 21 | x > 18)\). You can do this two ways:
    • Draw the graph where a is now 18 and b is still 25. The height is \(\frac{1}{(25-18)}=\frac{1}{7}\)
      So, \(P(x>21 | x>18)=(25-21)\left(\frac{1}{7}\right)=4 / 7\).
    • Use the formula: \(P(x>21 | x>18)=\frac{P(x>21 \cap x>18)}{P(x>18)}\)
      \(=\frac{P(x>21)}{P(x>18)}=\frac{(25-21)}{(25-18)}=\frac{4}{7}\).
  • 82.

    1. \(P(X>650)=\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}=0.125\).
    2. \(P(400<650)>

    84.

    1. \(X =\) the useful life of a particular car battery, measured in months.
    2. \(X\) is continuous.
    3. 40 months
    4. 360 months
    5. 0.4066
    6. 14.27

    86.

    1. \(X =\) the time (in years) after reaching age 60 that it takes an individual to retire
    2. \(X\) is continuous.
    3. five
    4. five
    5. Check student’s solution.
    6. 0.1353
    7. before
    8. 18.3

    88.

    a

    90.

    c

    92.

    Let \(X =\) the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean \(\lambda = 3\).
    Therefore, \((X=0)=\frac{3^{0} e^{-3}}{0 !}=e^{-3} \approx 0.0498\)

    Let \(T =\) duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, \(\mu = 7\) and the decay constant is \(m = \frac{1}{7}\). The cdf is \(P(T<t)>

    1. \(P(T<2)=1-1-e^{-\frac{2}{7}} \approx 0.2485\).
    2. \(P(T>15)=1-P(T<15)=1-\left(1-e^{-\frac{15}{7}}\right) \approx e^{-\frac{15}{7}} \approx 0.1173\).
    3. \(P(T>15 | T>10)=P(T>5)=1-\left(1-e^{-\frac{5}{7}}\right)=e^{-\frac{5}{7}} \approx 0.4895\).
    4. Let \(X = #\) of patients arriving during a half-hour period. Then \(X\) has the Poisson distribution with a mean of \(\frac{30}{7}\), \(X \sim \text{Poisson}(\frac{30}{7})\). Find \(P(X>8)=1-P(X \leq 8) \approx 0.0311\).