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%.
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?
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