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5.11: Chapter 5 Solution (Practice + Homework)

  • Page ID
    5570
  • 1.

    Uniform Distribution

    3.

    Normal Distribution

    5.

    \(P(6 < x < 7)\)

    7.

    one

    9.

    zero

    11.

    one

    13.

    0.625

    15.

    The probability is equal to the area from \(x = \frac{3}{2}\) to \(x = 4\) above the x-axis and up to \(f(x) = \frac{1}{3}\).

    17.

    It means that the value of \(x\) is just as likely to be any number between 1.5 and 4.5.

    19.

    \(1.5 \leq x \leq 4.5\)

    21.

    0.3333

    23.

    zero

    24.

    0.6

    26.

    \(b\) is 12, and it represents the highest value of \(x\).

    28.

    six

    30.

    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 5.38

    33.

    \(X =\) The age (in years) of cars in the staff parking lot

    35.

    0.5 to 9.5

    37.

    \(f(x) = \frac{1}{9}\) where x is between 0.5 and 9.5, inclusive.

    39.

    \(\mu = 5\)

    41.

    1. Check student’s solution.
    2. \(\frac{3.5}{7}\)

    43.

    1. Check student's solution.
    2. \(k = 7.25\)
    3. 7.25

    45.

    No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.

    47.

    five

    49.

    \(f(x)=0.2 e^{-0.2 x}\)

    51.

    0.5350

    53.

    6.02

    55.

    \(f(x) = 0.75e^{-0.75x}\)

    57.

    This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.75) on the y-axis and approaches the x-axis at the right edge of the graph. The decay parameter, m, equals 0.75.

    Figure 5.39

    59.

    0.4756

    61.

    The mean is larger. The mean is \(\frac{1}{m}=\frac{1}{0.75} \approx 1.33\), which is greater than 0.9242.

    63.

    continuous

    65.

    \(m = 0.000121\)

    67.

    1. Check student's solution
    2. \(P(x < 5,730) = 0.5001\)

    69.

    1. Check student's solution.
    2. \(k = 2947.73\)

    71.

    Age is a measurement, regardless of the accuracy used.

    73.

    1. Check student’s solution.
    2. \(f(x)=\frac{1}{8} \text { where } 1 \leq x \leq 9\)
    3. five
    4. 2.3
    5. \(\frac{15}{32}\)
    6. \(\frac{333}{800}\)
    7. \(\frac{2}{3}\)

    75.

    1. \(X\) represents the length of time a commuter must wait for a train to arrive on the Red Line.
    2. Graph the probability distribution.
    3. \(f(x)=\frac{1}{8} \text { where } 0 \leq x \leq 8\)
    4. four
    5. 2.31
    6. \(\frac{1}{8}\)
    7. \(\frac{1}{8}\)

    77.

    d

    78.

    b

    80.

    1. The probability density function of \(X\) is \(\frac{1}{25-16}=\frac{1}{9}\).
      \(P(X>19)=(25-19)\left(\frac{1}{9}\right)=\frac{6}{9}=\frac{2}{3}\).

      Figure 5.40

    2. \(P(19<22)>

      Figure 5.41

    3. 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, \((X=0)=\frac{3^{0} e^{-3}}{0 !}=e^{-3} \approx 0.0498\)

    NOTE

    You could let \(T =\) duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is \(\frac{1}{3}\) season. For the exponential, \(\mu = \frac{1}{3}\).
    Therefore, \(m = \frac{1}{\mu} = 3\) and \(T \sim E x p(3)\)

      The desired probability is \(P(T>1)=1-P(T<1)=1-\left(1-\mathrm{e}^{-3}\right)=\mathrm{e}^{-3} \approx 0.0498\).

      1. Let \(T =\) duration of time between no-hitters. We find \(P(T > 2|T > 1)\), and by the memoryless property this is simply \(P(T > 1)\), which we found to be 0.0498 in part a.
      2. Let \(X =\) the number of no-hitters is a season. Assume that \(X\) is Poisson with mean \(\lambda = 3\). Then \(P(X > 3) = 1 – P(X \leq 3) = 0.3528\).

      94.

      1. \(\frac{100}{9}=11.11\)
      2. \(P(X>10)=1-P(X \leq 10)=1-\text { Poissoncdf }(11.11,10) \approx 0.5532\).
      3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean \(\mu = 9\) and \(m = \frac{1}{9}\). The cumulative distribution function of \(X\) is \(P(X<x)>20)=1-P(X \leq 20)=1-\left(1-e^{-\frac{20}{9}}\right) \approx 0.1084\).

      NOTE

      We could also deduce that each person arriving has a \(8/9\) chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is \(\left(\frac{8}{9}\right)^{20} \approx 0.0948\). (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)

        96.

        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 \(P(X>8)=1-P(X \leq 8) \approx 0.0311\).