Question

in the 2000 cenus the so called long form received by one of every six households contained 52 quations ranging from your occupation and income all the way to weiher you had abathtab according to the us cenus bureau the mean competition time for the long form is 38 minutes assuming a standard deviation of 5 minutes and a simple of 50 persons who filled out the long form what is the probability thier average time for completion of the form was more than 45 minutes?

Answer 1

We can use the central limit theorem since we have a large enough sample size (n=50) and can assume that the distribution of completion times is approximately normal.

First, we need to find the z-score for a completion time of 45 minutes:

z = (45 - 38) / 5 = 1.4

Now we can use the z-score formula for sample means:

z = (x̄ - μ) / (σ / √n)

where x̄ is the sample mean, μ is the population mean (given as 38 minutes), σ is the population standard deviation (given as 5 minutes), and n is the sample size (given as 50).

Solving for x̄:

1.4 = (x̄ - 38) / (5 / √50)

1.4 * (5 / √50) = x̄ - 38

x̄ = 40.87

Therefore, the probability that the average completion time for the sample of 50 people was more than 45 minutes is the probability of getting a sample mean greater than 40.87.

Using a standard normal distribution table or calculator, we can find this probability:

P(Z > (45 - 38) / (5 / √50)) = P(Z > 3.54)

This probability is very small, approximately 0.0002. Therefore, we can conclude that it is very unlikely that the average completion time for the sample of 50 people was more than 45 minutes.