Q. There is a Chinese steel company that produces steel rods for local construction company. The plant manager knew from past statistics that the rods produced are normally distributed with a mean length of 300cm. The standard deviation is 8cm. If 22% of the sample means are more than specific length K, what is the value of K?
Solution: Given: u = 300cm, sigma = 8cm, n = 14, p(>K)= 0.22 Let K = be the length of the steel rod Since u (population) = u (sample) = 300, sigma (sample) = sigma /squareroot(14) = 2.138099 P(bar X > K) = [K - u(sample)]/2.138099 = Zvalue(0.22) (K - 300) / 2.138089935 = Zvalue(0.22) (K - 300) / 2.138089935 = 0.58 K = 2.138089935(0.58) + 300 K = 301.2401cm Therefore, if 22% of the sample means are more than a specific length K, then K is = 301.2401cm. Is my answer correct? . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
