SOLUTION: The weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 1 oz. From a batch of 1900 oranges, how many would be
Algebra ->
Probability-and-statistics
-> SOLUTION: The weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 1 oz. From a batch of 1900 oranges, how many would be
Log On
Question 1158828: The weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 1 oz. From a batch of 1900 oranges, how many would be expected to weight between 6 oz. and 8 oz., to the nearest whole number? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
for 6 oz z=-1.5/1
for 8 oz z=0.5/1
probability of z between those two values is 0.6247
multiply that by 1900, and one would expect 1186.84 or 1187 oranges to weigh in that interval
