Question: Normal Distribution Functions Norm.dist(x, mean, stdev, c) - calculates the normal distribution for a given value of x, where mean and stdev are the

Normal Distribution Functions Norm.dist(x, mean, stdev, c) - calculates the normal distributionfor a given value of x, where mean and stdev are the

Normal Distribution Functions Norm.dist(x, mean, stdev, c) - calculates the normal distribution for a given value of x, where mean and stdev are the descriptive statistics of the variable in question. Generally, the c argument will always be set to 1, as that calculates the cumulative probability up to the value x. Norm.inv(p, mean, stdev) - calculates the inverse normal distribution. Specifically, given a probability p, it will return the corresponding value of x in the distribution described by mean and stdev. Standard Normal Distribution Functions = x- - Standardize(x, mean, stdev) - returns the standard normal value z for a given value of x with a distribution described by mean and stdev. Recall that z Norm.s.dist(z) - calculates the standard normal distribution for a given value of z. The mean and stdev arguments are not needed, as they are 0 and 1, respectively. Norm.s.inv(p) - calculates the inverse standard normal distribution. Specifically, it returns a value for z, corresponding to a given probability p. 1 1 point earned The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23-7/8 inches and a standard deviation of 1/16 inch. 1a) Determine the mean of the difference between the width of the casing and the width of the door. 0.125 2 1 point earned 1a) Determine the standard deviation of the difference between the width of the casing and the width of the door. 0.1398 3 1 point earned 1b) What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch? 0.1867

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