Albert <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>... > 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?
I think you left out something. If the question concerns "22% of the sample means", a sample size should be provided. If the question were asked for 22% of the rods, your answer would be wrong. There are a few things that I think should be kept in mind: (1) Always make use of a sketch of the normal distribution for this type of problem. (2) Make sure that you distinguish between the Z values and the areas. (3) Work back from your answer to check. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
