Two stores sell watermelons. At the first store, the melons weigh an average of

wurmiana6d

wurmiana6d

Answered question

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?

Answer & Explanation

Froldigh

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 W=X+Y, the following properties hold for the mean, variance, and standard deviation:
E(W)=E(X)+E(Y)
V(W)=V(X)+V(Y) (if X and Y are indepedndent)
SD(W)=SD(X)2+SD(Y)2 (if X and Y are indepedndent)


a) Let W=XY - the difference in weights between the two stores:
E(W)=E(X)E(Y)=2218=4


b) Let W=XY - the difference in weights between the two stores:
SD(W)=SD(X)2+SD(Y)2=2.52+223.2016


c) The standardized score is the value x decreased by the mean and then divided by the standard deviation
z=xμσ=043.20161.25
Determine the corresponding probability using the normal table in the appendix:
P(XY>0)=P(Z1.25)=1P(Z<1.25)=10.1056=0.8944=89.44 %

 

Answers is
a) 4 pounds
b) 3.2016 pounds
c) P(XY>0)=0.8944=89.44 %

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