The weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 0.5 oz. What percentage of the oranges from the orchard weigh between 8 oz. and 9 oz., to the nearest tenth?
To find the percentage of oranges that weigh between 8 oz. and 9 oz., we need to calculate the z-scores for both weights and then look up the corresponding probabilities in the standard normal distribution table.
First, calculate the z-score for 8 oz:
\[z = \frac{X - \mu}{\sigma} = \frac{8 - 7.5}{0.5} = 1\]
Next, calculate the z-score for 9 oz:
\[z = \frac{X - \mu}{\sigma} = \frac{9 - 7.5}{0.5} = 3\]
Using the standard normal distribution table, we find that the probability of a z-score of 1 is approximately 0.8413 and the probability of a z-score of 3 is approximately 0.9987.
To find the percentage of oranges that weigh between 8 oz. and 9 oz., we need to subtract the probability of a z-score of 8 oz. from the probability of a z-score of 9 oz:
\[0.9987 - 0.8413 = 0.1574\]
Therefore, approximately 15.7% of the oranges from the orchard weigh between 8 oz. and 9 oz.