Question

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. Z .00 .01 .05 ооооо 0.0 0.1 0.2 0.3 0.4 0.5 0000 .0398 .0793 .1179 .1554 .1915 .0040 0438 .0832 .1217 .1591 1950 0199 .0596 0987 .1368 .1736 2088 on Au OOOO - 0.6 0.7 0.8 0.9 1.0 DO NO 2257 .2580 .2881 3159 .3413 .2291 2611 2910 3186 3438 .2422 2734 3032 3289 3531 1.2 1.3 .3643 .3849 4032 .4452 4332 .3665 13869 4049 4207 4345 -3749 3944 4115 4265 4394 1.5 1.6 1.7 1.8 1.9 2.0 4452 4554 4641 -4713 4772 4463 4564 .4649 4719 4778 4505 4599 4678 -4744 4789 2.1 2.2 2.3 2.4 2.5 4821 .4861 4893 4918 4938 4826 4864 4996 4920 4940 4842 4878 4906 4929 4946 2.6 2.7 2.8 2.9 3.0 4953 4965 .4974 4981 4987 4955 4966 4975 4982 4987 4960 4970 4978 4984 -4989 . 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 O. 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. %. Unlimited Attempts Remain​

Answer 1

Answer:2.07 is the answer as of e.2020

Step-by-Step: 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%