2.3 Probability Distributions

Translating Word to Symbols



  1. In the world, 10% of people are left-handed. Fifty people are randomly selected, translate the following into probability statements.
    Words Probability Statement
    Probability of at least 21 out of 50

    \(P(x \geq 21)\)
    Probability of less than 5 out of 50

    \(P(x<5)\)
    Probability of no more than 9 out of 50

    \(P(x \leq 9)\)
    Probability of more than 3 out of 50

    \(P(x>3)\)
    Probability of at least 2 out of 50

    \(P(x \geq 2)\)
    Probability of 6 out of 50

    \(P(x=6)\)
    Probability of less than 4 out of 50

    \(P(x<4)\)

    2. Discrete Probability Distributions

  2. Human blood comes in different types. Each person has a specific ABO type (A, B, AB, or O) and Rh factor (positive or negative). Hence, if you are O+, your ABO type is O and your Rh factor is positive. A probability model for blood types in the United States is given:
    Blood Type A+ A- B+ B- AB+ AB- O+ O-
    Probability 0.357 0.063 0.085 0.015 0.034 0.006 0.374 0.066
    1. Does this problem meet all the requirements for a binomial probability distribution? Explain how you know. Yes
    2. What is the probability that a randomly chosen person has blood type A?

      3. \(P(A+\text { or } A-)=P(A+)+P(A-)=0.357+0.063=0.420\)

      Hence, roughly 42% of residents in the U.S. have ABO-type A blood.

    3. What is the probability that two randomly chosen U.S. residents both have type A blood?

      4. \(\text { P(Person 1 is type A and Person 2 is type A) = }(0.420)(0.420)=0.1764\)

      Hence, there is roughly an 18% chance that two randomly selected U.S. residents will both have type A blood.

    4. What is the probability that two U.S. residents are randomly chosen and neither has Type A blood?

      \((0.58)(0.58)=0.3364\)

    5. What is the probability that at least one of the two people has type A blood?

      6. \( \text {P(at least one has A)}=1-P(\text { neither A } )=1-0.3364=0.6636\)


    In a probability distribution,
    • \(\sum P(x)=1\): the sum of the probabilities must equal 1.
    • \(0 \leq P(x) \leq 1\): all the probabilities are less than 1 and positive.
  3. A cell phone service has the following distribution of cell phones associated with each friends and family account.

    (The symbol 0+ denotes a positive probability value that is very small. We are not willing to say this is an impossible event. Use P(x=6) = 0 in your calculations.)

    x values Probability P(x)
    1 .02
    2 .13
    3 .32
    4 .39
    5 .14
    6 0+

    1. Does this satisfy both rules for probabiliy distributions?

      Yes. \({\textstyle\sum_{}^{}}P(x)=1\) and \(0 < P(x) < 1\)

    2. If one account is randomly selected, find the following probabilities for the number of phones associated with the account:
      1. Probability of exactly 3 phones:

        P(x = 3) = .32

      2. Probability of 4 or more phones:

        P(x ≥ 4) = P(x = 4) + P(x = 5) + P(x = 6) = .39 + .14 + 0 = .53

      3. Probability of at least 2 phones:

        P(x ≥ 2) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = .13 + .32 + .39 + .14 + 0 = .98

        Or use the complement rule: 1 – P(x = 1) = 1 - .02 = .98

      4. Probability of at most 3 phones:

        P(x ≤ 3) = .02 + .13 + .32 = .47

      5. Probability of no more than 2 phones:

        P(x ≤ 2) = .02 + .13 = .15

      6. Probability of between 2 and 5 phones:

        P(2 < x < 5) = P(x = 3) + P(x = 4) = .32 + .39 = .71

      7. Probability of between 2 and 5 phones, inclusive:

        P(2 ≤ x ≤ 5) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) = .13 + .32 + .39 + .14 = .98

    3. MEAN (EXPECTED VALUE): \(\mu=E(x)=\sum[x * P(x)]\)

      \(\mu=[1 * .02]+[2 * .13]+[3 * .32]+[4 * .39]+[5 * .14]+[6 * 0]=3.5\)

    4. STANDARD DEVIATION:

      \(\sigma=\sqrt{\sigma^{2}}=\sqrt{.91}=.95\)

    5. UNUSUAL VALUES:

      Min usual value: \(\mu-2 \sigma = \) \(3.5-2(.95)=1.6\)

      Max usual value: \(\mu+2 \sigma = \) \(3.5+2(.95)=5.4\)

      Would selecting an account with 1 phone be unusual? Yes, 1 is less than the min usual value.

      Would selecting an account with 2 phones be unusual? No, 2 is within the usual range.

      Would selecting an account with 5 phones be unusual? No, 5 is within the usual range.

  4. The following table shows the distribution of family sizes in a population. Use the data in the table below to answer the questions.

    y 1 2 3 4 5 6 7 8
    p(y) 0.15

    0.23

    		 0.19 0.23 0.12 0.05 0.02 0.01


    1. Does this problem meet all the requirements for a binomial probability distribution? Explain how you know. Yes
    2. What is the probability that a randomly selected family will consist of exactly two people?

      3. \(P(y = 2)=1-0.15-0.19-0.23-0.12-0.05-0.02-0.01=0.23\)

    3. What is the probability that a randomly selected family will consist of at least two people?

      4. \(P(y \geq 2)=1-P(y\lt2)=1-p(1)=0.85\)

    4. What is the probability that a randomly selected family will have two to four members?

      5. \(\mathrm{P}(2 \leq \mathrm{y} \leq 4)=\mathrm{P}(\mathrm{y}=2 \text { or } \mathrm{y}=3 \text { or } \mathrm{y}=4)=\mathrm{p}(2)+\mathrm{p}(3)+\mathrm{p}(4)=0.65\)

    5. What is the mean family size?

      6. \(\mu=3.22\)

    6. What is the standard deviation for family size?

      7. \(\sigma \approx \sqrt{2.512} \approx 1.58\)

    7. Would selecting a family with 8 members be unusual?

      8. \(\text { The minimum usual value is } \mu-2 \sigma=3.22-2(1.58) \text { which is } 0.06\)

      \(\text { The maximum usual value is } \mu+2 \sigma=3.22+2(1.58) \text { which is } 6.38.\)

      So, yes families with 7 or 8 members are unusual.