6. (15 points) Let X denote the length of the side of a manufacturer part that h

Question

6. (15 points) Let X denote the length of the side of a manufacturer part that has a square shape (measured in centimeters).Based on historical data, the manufacturer knows that X N (2,0.15). What is the mean and variance of the perimeter of this part (Hint: The perimeter of a square equals the sum of the lengths of its four sides) Determine the probability that the perimeter exceeds 8.5 centimeters If an engineer picks a sample of 34 parts, what is the probability that the average of perimeters of these 34 parts () would be less than 8? a. b. c.

Explanation / Answer

Here X ~ N (2,0.15)

(a) So perimeter P = 4X

so E(P) = E[4X] = 4 * E[X] = 4 * 2 = 8

Var(P) = 4 * Var(X) = 4 * 0.152 = 0.09

(b) Here

standard deviation of perimeter, STD (P) = sqrt (0.09) = 0.3

so,

Pr(X > 8.5 ; 8 ; 0.30) = 1 - Pr(X < 8.5 ; 8 ; 0.30)

Z = (8.5 - 8)/0.30 = 1.6667

so,

Pr(X > 8.5 ; 8 ; 0.30) = 1 - Pr(X < 8.5 ; 8 ; 0.30) = 1 - Pr(Z < 1.6667) = 1 - 0.9522 = 0.0478

(c) Here the mean of 34 samples are 8 as per normal distribution as the distribution is symmetric so there would be 50% probability that the average of the peremeter is less than 8 cm.

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