Question 1184550: A manufacturer knows that their items have a normally distributed length, with a mean of 5.5 inches, and standard deviation of 0.7 inches.
If 5 items are chosen at random, what is the probability that their mean length is less than 4.7 inches? (Give answer to 4 decimal places.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the sample size is 5.
if you follow this guide as to when to use the t-score or when to use the z-score, you will find that the use of the t-score is indicated in this problem.
https://math.stackexchange.com/questions/1817980/how-to-know-when-to-use-t-value-or-z-value
i'll do both to show you the difference.
the population mean is 5.5 inches.
the population standard deviation is .7 inches.
the standard error = population standard deviation / square root of sample size = .7 / sqrt(5) = .31305.
the t-score is equal to (x - m) / s = (4.7 - 5.5) / .31305 = -2.5555.
the test t-score with 4 degrees of freedom give you an area of .0314687 to the left of that z-score.
your probability that the mean length is less than 4.7 inches is equal to .0315 rounded to 4 decimal places.
the z-score is also equal to -2.5555.
the probability of getting a z-score less than -2.5555 is equal to .0053.
here's a reference on the relative shape of the t-score, depending on the degrees of freedom.
https://www.statisticshowto.com/t-statistic/
the t-score begins to look more like the z-score as the sample size goes up.
at the lower end of sample sizes, the difference is more pronounced.
your solution in this problem is that the probability of getting a score less than 4.7 is equal to .0315 rounded to 4 decimal places.
this assumes use of the t-score.
if the use of the z-score is assumed, than the probability is .0053.
i would go with the assumption of t-score, unless otherwise instructed.
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