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20 June, 17:13

A university wants to estimate the average distance that commuter students travel to get to class with an error of ±3 miles and 90 percent confidence. What sample size would be needed, assuming that travel distances are normally distributed with a range of X = 0 to X = 50 miles, using the Empirical Rule μ ± 3σ to estimate σ.

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  1. 20 June, 17:30
    0
    n = 21

    Step-by-step explanation:

    Solution:-

    - Let denote a random variable "X" : average distance that commuter students travel to get to class.

    - The population is given to be normally distributed, such that:

    Range X: [ 0, 50 ] miles

    - We will use the given range coupled with the empirical rule for normal distribution to determine the mean (u) and standard deviation of population (σ):

    P (μ - 3σ < X < μ + 3σ) = 0.997 ... (Empirical Rule)

    - According to the standardized results for Z-table:

    P (-3 < Z < 3) = 0.997

    So, P (Z ≤ 3) = 1 - (1 - 0.997) / 2 = 0.9985

    P (Z ≥ - 3) = 1 - (1 - 0.997) / 2 = 0.9985

    - The standardized values for the given data can now be determined:

    P (X ≥ μ - 3σ) = P (Z ≥ - 3) = 0.9985

    X ≥ μ - 3σ = Upper limit - 0.9985 * (Range)

    X ≥ μ - 3σ = 50 - 0.9985 * (50)

    μ - 3σ = 0.075 ... Eq1

    P (X ≤ μ + 3σ) = P (Z ≤ 3) = 0.9985

    X ≤ μ + 3σ = Lower limit + 0.9985 * (Range)

    X ≤ μ + 3σ = 0 + 0.9985 * (50)

    μ + 3σ = 49.925 ... Eq2

    - Solve the Eq1 and Eq2 simultaneously:

    2μ = 50, μ = 25 miles

    3σ = 24.925

    σ = 8.30833

    - Hence, the normal distribution parameters are:

    X ~ N (μ, σ^2)

    X ~ N (25, 8.308^2)

    - The standard error in estimation of average distance that commuter students travel to get to class is E = ±3 miles for the confidence level of 90%.

    - The Z-critical value for confidence level of 90%, Z-critical = 1.645

    - The standard error estimation statistics is given by the following relation with "n" sample size.

    E = Z-critical*σ / √n

    n = [ Z-critical*σ / E ]^2

    - Plug in the values:

    n = [ 1.645*8.308/3]^2

    n = 20.75306 ≈ 21

    Answer: The sample size needed to estimate average distance that commuter students travel to get to class with error of ±3 miles and 90 percent confidence, is n = 21.
  2. 20 June, 17:39
    0
    A university wants to estimate the average distance that commuter students travel to get to class with an error of ±3 miles and 90 percent confidence. What sample size would be needed, assuming that travel distances are normally distributed with a range of X = 0 to X = 50 miles, using the Empirical Rule μ ± 3σ to estimate σ.

    The required sample size, n = (zσ/E) ² = 21.0

    Step-by-step explanation:

    The estimated σ here = (range) / 6 = (50/6) = 8.33

    In the case of 90 %, CI value of z = 1.64

    standard deviation, σ = 8.33

    margin of error E = 3

    The required sample size, n = (zσ/E) ² = 21.0
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