The weight of oranges growing in an orchard is normally distributed with a mean weight of 3.5 oz. and a standard deviation of 1 oz. What is the probability that a randomly selected orange from the orchard weighs less than 2 oz., to the nearest thousandth?

Answer 1

To solve this problem, we need to standardize the given weight using the z-score formula:

z = (X - μ) / σ

Where:
X = 2 oz. (weight of the orange we want to find the probability for)
μ = 3.5 oz. (mean weight of the oranges in the orchard)
σ = 1 oz. (standard deviation of the weights)

Plugging in the values, we get:

z = (2 - 3.5) / 1
z = -1.5

Now, we need to find the probability of a z-score less than -1.5. We can use a standard normal distribution table or calculator to find this probability.

Using a standard normal distribution table, the probability of a z-score less than -1.5 is approximately 0.0668.

Therefore, the probability that a randomly selected orange from the orchard weighs less than 2 oz. is approximately 0.0668 (or 6.68% to the nearest thousandth).