The weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz. and a standard deviation of 1.5 oz. What percentage of the oranges from the orchard weigh less than 7 oz., to the nearest tenth?
To solve this problem, we need to find the z-score of 7 oz. using the formula:
z = (X - μ) / σ
Where:
X = 7 oz.
μ = 8 oz. (mean weight)
σ = 1.5 oz. (standard deviation)
Plugging in the values, we get:
z = (7 - 8) / 1.5
z = -0.67
Next, we can look up the z-score in a standard normal distribution table or use a calculator to find that the percentage of oranges weighing less than 7 oz. is approximately 25.0% to the nearest tenth.