Hugo averages 58 words per minute on a typing test with a standard deviation of 11 words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then, X∼N(58,11). Suppose Hugo types 74 words per minute in a typing test on Wednesday. The z-score when x=74 is ________. This z-score tells you that x=74 is ________ standard deviations to the ________ (right/left) of the mean, ________.

Answer:The number of words that Hugo wrote in the test of Wednesday (74) is at a distance of 1.4545 standard deviations (11) to the right of his mean (58)Step-by-step explanation:A normal random variable with mean Mu = 58 and standard deviation sd = 11 is standardized with the transformation (z-score):Z = (X - Mu) / sd = (X - 58) / 11For a value of 74 for X, Z = (74 - 58) / 11 = 1.4545 > 0.The z-score when X = 74 is 1.4545 > 0This means that the number of words that Hugo wrote in the test of Wednesday (74) is at a distance of 1.4545 standard deviations (11) to the right of his average (58)