wurmiana6d
2021-11-21
Two stores sell watermelons. At the first store, the melons weigh an average of 22 pounds, with a standard deviation of 2.5 pounds. At the second store, the melons are smaller, with a mean of 18 pounds and a standard deviation of 2 pounds. You select a melon at random at each store. a) What's the mean difference in weights of the melons? b) What's the standard deviation of the difference in weights? c) If a Normal model can be used to describe the difference in weights, what 's the probability that the melon you got at the first store is heavier?
Froldigh
Beginner2021-11-22Added 17 answers
The linear combination of the anticipated values is the expected value of a linear combination.
A linear combination's variance is equal to its linear combination times the square of its variance coefficients (constant terms have no effect on the variance).
For the sum , the following properties hold for the mean, variance, and standard deviation:
(if X and Y are indepedndent)
(if X and Y are indepedndent)
a) Let - the difference in weights between the two stores:
b) Let - the difference in weights between the two stores:
c) The standardized score is the value x decreased by the mean and then divided by the standard deviation
Determine the corresponding probability using the normal table in the appendix:
%
Answers is
a) 4 pounds
b) 3.2016 pounds
c) %
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