Respuesta :
Answer:
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
Answer:
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         Â
          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 Â
- The standard error estimation statistics is given by the following relation with "n" sample size.
           E = Z-critical*σ /√n  Â
           n = [ Z-critical*σ /E ]^2
- Plug in the values:
           n = [ 1.645*8.308/3]^2  Â
           n = 20.75306 ≈ 21 Â
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.   Â
