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.

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

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  

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.      

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation: