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.

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.

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