Let random variable w = width in inches across the rump of a randomly selected h

Question

Let random variable w = width in inches across the rump of a randomly selected horse a) The usual length measure as it applies to horses is the "hand, " which is by definition equal to 4". Define random variable h = width in hands across the rump of a randomly selected horse What are the mean and standard deviation of h? b) Suppose horses are randomly chosen for a particular Roman chariot. Define the random variable c = h1 + h2 to be the rump width of two randomly selected horses. What are the mean and standard deviation of c? c) Suppose that in the original chariot design a 16 inch separation of the horses is specified so that the horses have room to avoid each other. This leads to random variable a =16+ w1 + w2. Describe how this addition of 16 inches of "wiggle room" would change the mean and standard deviation in part (b). Do not recalculate the mean and standard deviation.

Explanation / Answer

Solution

Back-up Theory

H = width (in hands) across the rump of a horse. Then, H ~ N(µ, 2)………………………..(1)

If X ~ N(µ1, 12) and Y ~ N(µ2, 22), then X + Y ~ N(µ1 + µ2 , 12 + 22), when X and Y are independent…………………………………………………………………………………….(2)

E(aX + b) = aE(X) + b ………………………………………………………………………..(3)

V(aX + b) = a2V(X) ………………………………………………………………………..(4)

Now, to work out solution,

Part (a)

[vide (1) under Back-up Theory], mean of H is µ and standard deviation of H is . ANSWER

Part (b)

If C = H1 + H2, then [vide (2) under Back-up Theory], C ~ N(2µ, 22). So,

Mean of C = 2µ and standard deviation = (2) ANSWER

Part (c)

If A = 16 + H1 + H1, [vide (3) and (4) under Back-up Theory], mean of A is 16 + 2µ and standard deviation of A is (2). ANSWER

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