Question 1182714: The average amount of money spent for lunch per person in the college cafeteria is $6.45 and the standard deviation is $2.55.
Suppose that 44 randomly selected lunch patrons are observed.
Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.
b. What is the distribution of x?
x~N(6.45,_________________)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! here are the basic rules for when to use t-score or z-score.
https://www.algebra.com/statistics/Central-limit-theorem/change-this-name2886.lesson?content_action=show_prod
your population mean is 6.45
your population standard deviation is 2.55.
your sample size is 44.
z-score is indicated because you are using population standard deviation and sample size is greater than 30.
you want to compare the sample mean to the population mean.
the z-score formula is (x - m) / s
in this case, x is the sample mean, m is the population mean, s is the standard error.
standard error = standard deviation / sqrt(sample size0 = 2.55 / sqrt(44) = .3844 rounded to 4 decimal places.
z-score formula becomes z = (x - 6.45) / .3844
solve for x to get:
x = .3844 * z + 6.45
since you don't have a z-score, you can't find x.
the best you can do is find the critical z-scores, and hence, the critical raw scores.
in order to do that you need to to know the confidence level.
if you know that, you can find the critical z-scores and, from that, the critical x-scores.
we'll compare two confidence levels.
the first is at 99% confidence level.
the second is at 90% confidence evel.
with a 99% two-tail confidence level, the critical z-scores are plus or minus 2.5758 rounded to 4 decimal places.
the critical raw scores are found by using the z-score formula.
you get x = .3844 * 2.5758 + 6.45 = 7.4401 for the high score.
you get x = .3844 * -2.5758 + 6.45 = 5.4599 for the low score.
the 99% two-tail confidence level tells you that 99% of your samples of size 44 will have the sample mean between 5.4599 and 7.4401.
that's your 99% confidence interval.
with a 95% two-tail confidence level, the critical z-scores are plus or minus 1.9600 rounded to 4 decimal places.
the critical raw scores are found by using the z-score formula.
you get x = .3844 * 1.9600 + 6.45 = 7.2034 for the high score.
you get x = .3844 * -1.9600 + 6.45 = 5.6966 for the low score.
the 95% two-tail confidence level tells you that 95% of your samples of size 44 will have the sample mean between 5.6966 and 7.2034.
that's your 95% confidence interval.
the higher the confidence level, the wider the confidence interval when you are dealing with the same population mean and population standard deviation and same sample size.
visually this would look like this:
99% confidence interval.
95% confidence interval.
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