The number of flaws per square yard in a type of carpet material varies with mean 1.7 flaws per square yard and standard deviati
on 0.8 flaws per square yard. This population distribution cannot be normal, because a count takes only whole-number values. An inspector studies 173 square yards of the material, records the number of flaws found in each square yard, and calculates x, the mean number of flaws per square yard inspected. Use the central limit theorem to find the approximate probability that the mean number of flaws exceeds 1.8 per square yard. (Round your answer to four decimal places.)
Let x be a random variable representing the mean number of flaws per square yard inspected P(x > 1.8) = P(z > (1.8 - 1.7)/0.8) = P(z > 0.125) = P(z < 0.125) = 0.5497
Let p (x) = 4x^4-13x^2-2x The zero of x-2 is 2 putting x = 2 in p (x), we get, p (2) = 4×2^4-13×2^2-2×2 = 64 - 52 - 4 = 64 - 56 = 8 Therefore, remainder = 8
