Find z for 90th percentile (1.28)
xbar = mean weight we want to find
z = (xbar - mu)/(s/sqrt(n)) = (xbar - 511)/(29/sqrt(25))
1.28 = (xbar - 511)/(29/sqrt(25))
1.28 = (xbar - 511)/5,8
xbar = 1.28 * 5.8 + 511 = 518.424 ~ 518 g
Ben Y.
asked 07/21/22A particular fruit's weights are normally distributed, with a mean of 511 grams and a standard deviation of 29 grams.
If you pick 25 fruits at random, then 10% of the time, their mean weight will be greater than how many grams?
Give your answer to the nearest gram.
Find z for 90th percentile (1.28)
xbar = mean weight we want to find
z = (xbar - mu)/(s/sqrt(n)) = (xbar - 511)/(29/sqrt(25))
1.28 = (xbar - 511)/(29/sqrt(25))
1.28 = (xbar - 511)/5,8
xbar = 1.28 * 5.8 + 511 = 518.424 ~ 518 g
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