Math Problem Statement

Develop a discrete probability distribution for x = Age.

What is the probability that a student's age is more than 20?

What is the mean age?

What is the standard deviation of the ages?

Suppose that we consider the distribution of ages to be approximately bell-shaped. We would expect 95% of the ages to lie between and years. (Remember to use non-rounded values in your calculations and only round your final answers.)

Solution

To approach this problem, let’s outline the steps:

  1. Discrete Probability Distribution: We need to know the specific age values and their corresponding probabilities. If you have any specific age data, we can use that. For now, I will assume we have data about student ages.

    Let's suppose we have the following age distribution of students:

    Age (x)Probability P(x)
    180.20
    190.25
    200.30
    210.15
    220.10
  2. Find the probability that a student's age is more than 20:

    To find P(x>20)P(x > 20), we sum the probabilities for ages greater than 20:

    = 0.15 + 0.10 = 0.25$$
  3. Mean (Expected Value) of Age:

    The mean (expected value) μ\mu of the distribution can be found using the formula: μ=(xP(x))\mu = \sum (x \cdot P(x)) Substituting the values from the table:

    = 3.6 + 4.75 + 6.0 + 3.15 + 2.2 = 19.7$$ So, the mean age $$\mu = 19.7$$ years.
  4. Standard Deviation of Ages:

    The standard deviation σ\sigma is calculated using the formula: σ=(xμ)2P(x)\sigma = \sqrt{\sum (x - \mu)^2 \cdot P(x)} First, calculate (xμ)2P(x)(x - \mu)^2 \cdot P(x) for each value of xx:

    Age (x)(xμ)(x - \mu)(xμ)2(x - \mu)^2(xμ)2P(x)(x - \mu)^2 \cdot P(x)
    181819.7=1.718 - 19.7 = -1.72.892.892.89×0.20=0.5782.89 \times 0.20 = 0.578
    191919.7=0.719 - 19.7 = -0.70.490.490.49×0.25=0.12250.49 \times 0.25 = 0.1225
    202019.7=0.320 - 19.7 = 0.30.090.090.09×0.30=0.0270.09 \times 0.30 = 0.027
    212119.7=1.321 - 19.7 = 1.31.691.691.69×0.15=0.25351.69 \times 0.15 = 0.2535
    222219.7=2.322 - 19.7 = 2.35.295.295.29×0.10=0.5295.29 \times 0.10 = 0.529

    Now, sum up the last column: (xμ)2P(x)=0.578+0.1225+0.027+0.2535+0.529=1.51\sum (x - \mu)^2 \cdot P(x) = 0.578 + 0.1225 + 0.027 + 0.2535 + 0.529 = 1.51 Finally, the standard deviation is: σ=1.511.23\sigma = \sqrt{1.51} \approx 1.23

  5. 95% Confidence Interval (Assuming Bell-Shaped Distribution):

    If the distribution is approximately bell-shaped (normal distribution), 95% of the data falls within μ±2σ\mu \pm 2\sigma.

    Lower Bound=μ2σ=19.72(1.23)=19.72.46=17.24\text{Lower Bound} = \mu - 2\sigma = 19.7 - 2(1.23) = 19.7 - 2.46 = 17.24 Upper Bound=μ+2σ=19.7+2(1.23)=19.7+2.46=22.16\text{Upper Bound} = \mu + 2\sigma = 19.7 + 2(1.23) = 19.7 + 2.46 = 22.16

Thus, we expect 95% of student ages to lie between 17.24 years and 22.16 years.


Summary:

  1. The probability that a student's age is more than 20 is 0.25.
  2. The mean age is 19.7 years.
  3. The standard deviation is approximately 1.23 years.
  4. We expect 95% of the ages to lie between 17.24 and 22.16 years.

Would you like more details or have any questions?

Here are some related questions:

  1. How would changing the probabilities of certain age groups affect the mean age?
  2. How would you calculate the probability for a student's age being between 19 and 21?
  3. How is the mean age impacted if the majority of students were slightly older?
  4. How would you compute the median of this age distribution?
  5. Can the skewness of the distribution impact the mean or standard deviation?

Tip: The standard deviation measures how spread out values are around the mean. The larger the standard deviation, the wider the spread of values.

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Math Problem Analysis

Mathematical Concepts

Probability Distribution
Expected Value
Standard Deviation
Normal Distribution

Formulas

Mean (Expected Value) formula: μ = Σ(x * P(x))
Standard Deviation formula: σ = √Σ((x - μ)^2 * P(x))
Probability P(x > 20): P(21) + P(22)
Confidence Interval for normal distribution: μ ± 2σ

Theorems

Law of Large Numbers
Empirical Rule (68-95-99.7 Rule) for normal distribution

Suitable Grade Level

Grade 11-12, College Introductory