Apple weights in an orchard are normally distributed. from a sample farmer fred determines the mean weight of a box of apples to be 270 oz. with a standard deviation of 10 oz. he wonders what percent of the apple boxes he has grouped for sale will have a weight less than 255 oz. carl calculates the z-score corresponding to the weight 255 oz. = - using the table (column. 00), carl sees the area associated with this z-score is 0. carl rounds this value to the nearest hundredth or 0. now, carl decides to find the percent associated with boxes less than 255 oz. so he rounds to hundredths and subtracts 0.50 - 0. = %.

Question

Mathematics

Morgan Lynch

25 April, 08:49

Apple weights in an orchard are normally distributed. from a sample farmer fred determines the mean weight of a box of apples to be 270 oz. with a standard deviation of 10 oz. he wonders what percent of the apple boxes he has grouped for sale will have a weight less than 255 oz. carl calculates the z-score corresponding to the weight 255 oz. = - using the table (column. 00), carl sees the area associated with this z-score is 0. carl rounds this value to the nearest hundredth or 0. now, carl decides to find the percent associated with boxes less than 255 oz. so he rounds to hundredths and subtracts 0.50 - 0. = %.

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Byron Noble · Answer 1

Mean = 270 Standard deviation = 10 x = 255 Formula for z-score, z = (x - mean) / SD z = (255 - 270) / 10 = > z = - 15 / 10 = > z = - 1.5 So by referring to z-table, - 1.5 correlates to 0.0668 that implies to 0.07 So 7% of the boxes of Apples weight less than 255oz. The percentage of boxes is in the range of 255 oz and 270 oz, Now calculating the requiring percentage 50% - 7% = 43%