A manager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium shells. Past data indicate that if the machine is functioning properly, the length of the shells produced by this machine can be modeled as being normally distributed with a mean of 118 centimeters and a standard deviation of 6.3 centimeters. Suppose 10 shells produced by this machine are randomly selected. What is the probability that the average length of these 10 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

Answer:0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".Step-by-step explanation:To solve this question, we need to understand the normal probability distribution and the central limit theorem.Normal Probability Distribution:Problems of normal distributions can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.Central Limit TheoremThe Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters. This means that [tex]\mu = 118, \sigma = 8[/tex]Sample of 16 shellsThis means that [tex]n = 16, s = \frac{8}{\sqrt{16}} = 2[/tex]What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?" This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.X = 120[tex]Z = \frac{X - \mu}{\sigma}[/tex]By the Central Limit Theorem[tex]Z = \frac{X - \mu}{s}[/tex][tex]Z = \frac{120 - 118}{2}[/tex][tex]Z = 1[/tex][tex]Z = 1[/tex] has a pvalue of 0.841X = 116[tex]Z = \frac{X - \mu}{s}[/tex][tex]Z = \frac{116 - 118}{2}[/tex][tex]Z = -1[/tex][tex]Z = -1[/tex] has a pvalue of 0.1590.841 - 0.159 = 0.6820.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".