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?

Question

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&#039;s the mean difference in weights of the melons? b) What&#039;s the standard deviation of the difference in weights? c) If a Normal model can be used to describe the difference in weights, what &#039;s the probability that the melon you got at the first store is heavier?

Froldigh · Accepted Answer

The linear combination of the anticipated values is the expected value of a linear combination.A linear combination&#039;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&amp;nbsp;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)&amp;nbsp;(if X and Y are indepedndent)SD(W)=SD(X)2+SD(Y)2&amp;nbsp;(if X and Y are indepedndent)a) Let&amp;nbsp;W=X−Y&amp;nbsp;- the difference in weights between the two stores:E(W)=E(X)−E(Y)=22−18=4b) Let&amp;nbsp;W=X−Y&amp;nbsp;- the difference in weights between the two stores:SD(W)=SD(X)2+SD(Y)2=2.52+22≈3.2016c) The standardized score is the value x decreased by the mean and then divided by the standard deviationz=x−μσ=0−43.2016≈−1.25Determine the corresponding probability using the normal table in the appendix:P(X−Y&amp;gt;0)=P(Z≻1.25)=1−P(Z&amp;lt;−1.25)=1−0.1056=0.8944=89.44&amp;nbsp;%&amp;nbsp;Answers isa) 4 poundsb) 3.2016 poundsc)&amp;nbsp;P(X−Y&amp;gt;0)=0.8944=89.44&amp;nbsp;%

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

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