The weight of oranges growing in an orchard is normally distributed with a mean weight of 4 oz. and a standard deviation of oz What percentage of the oranges from the orchard weigh between 4 oz. and 5 oz., to the nearest tenth?
To find the percentage of oranges that weigh between 4 oz. and 5 oz., we need to calculate the z-scores for both weights and then use a standard normal distribution table to find the area under the curve between those two z-scores.
First, calculate the z-scores for 4 oz. and 5 oz.
Z(4 oz) = (4 - 4) / 1 = 0
Z(5 oz) = (5 - 4) / 1 = 1
Next, we look up the z-scores in a standard normal distribution table. The area to the left of z = 0 is 0.5000, and the area to the left of z = 1 is 0.8413.
To find the area between the z-scores of 0 and 1, we subtract the cumulative probability at z = 0 from the cumulative probability at z = 1: 0.8413 - 0.5000 = 0.3413.
Therefore, approximately 34.13% of the oranges from the orchard weigh between 4 oz. and 5 oz.