A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. If the population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch, the probability that Claude's sample has a mean between 119.985 and 120.0125 inches is

Answer:There is a 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.Step-by-step explanation:The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.In this problem, we have that:The population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch. This means that [tex]\mu = 120, \sigma = 0.05[/tex].Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. This means that [tex]n = 100, s = \frac{\sigma}{\sqrt{n}} = \frac{0.05}{\sqrt{100}} = 0.005[/tex].The probability that Claude's sample has a mean between 119.985 and 120.0125 inches isWe are working with the sample mean, so we use the standard deviation of the sample, that is, [tex]s[/tex] instead of [tex]\sigma[/tex] in the z score formula.This probability is the pvalue of Z when [tex]X = 120.0125[/tex] subtracted by the pvalue of Z when [tex]X = 119.985[/tex].X = 120.0125[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{120.0125 - 120}{0.005}[/tex][tex]Z = 2.5[/tex][tex]Z = 2.5[/tex] has a pvalue of 0.9938.X = 119.985[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{119.985 - 120}{0.005}[/tex][tex]Z = -3[/tex][tex]Z = -3[/tex] has a pvalue of 0.0014.So there is a 0.9938 - 0.0014 = 0.9924 = 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.