in the 2000 cenus the so called long form received by one of every households contained 52 quations ranging form your occupation and income all the way to weiher you had a bathtub according to the us cenus bureau the mean completion time for the long form is 38 minutes assuming a standard deviation of 5 minutes and a simple random sample 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?
To solve this problem, we will use the Central Limit Theorem.
The mean completion time for the long form is given to be 38 minutes with a standard deviation of 5 minutes.
First, we need to find the standard error (SE) of the sample mean. The formula for the standard error is given by:
SE = standard deviation / √(sample size)
SE = 5 / √(50) ≈ 0.707
Next, we need to calculate the z-score for the given average completion time of 45 minutes. The z-score formula is:
z = (X - μ) / SE
Where X is the given value, μ is the mean, and SE is the standard error.
z = (45 - 38) / 0.707 ≈ 9.89
Now, we need to find the probability that the average completion time is more than 45 minutes, which is equivalent to finding the area to the right of the z-score on a standard normal distribution.
Using a standard normal distribution table or calculator, we find that the probability associated with a z-score of 9.89 is approximately 1 (or 100%).
Therefore, the probability that their average completion time for the form was more than 45 minutes is approximately 1 or 100%.
To solve this problem, we will use the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases.
Given:
- Mean completion time for the long form = 38 minutes
- Standard deviation = 5 minutes
- Sample size = 50
We need to find the probability that the average time for completion of the form was more than 45 minutes.
First, let's calculate the standard error (SE) of the sample mean:
SE = standard deviation / √sample size
SE = 5 / √50
SE ≈ 0.707 (rounded to three decimal places)
Next, we convert the given average time of 45 minutes to a z-score using the formula:
z = (x - μ) / SE
z = (45 - 38) / 0.707
z ≈ 9.899 (rounded to three decimal places)
Now, we can find the probability using the z-score table or a calculator:
P(Z > 9.899) (where Z is the standard normal distribution)
Looking at the z-score table, we can see that the probability for a z-score of 9.899 is essentially zero.
Therefore, the probability that their average time for completion of the form was more than 45 minutes is almost zero.