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 What is the probability that a randomly selected orange from the orchard weighs more than 10 oz., to the nearest thousandth?
To solve this problem, we need to calculate the z-score of the weight of 10 oz.
First, we calculate the z-score:
z = (X - μ) / σ
z = (10 - 8) / 1.5
z = 2 / 1.5
z = 1.3333
Next, we use a standard normal distribution table or a calculator to find the probability that a randomly selected orange weighs more than 10 oz. This corresponds to the area under the normal curve to the right of the z-score of 1.3333.
Using a standard normal distribution table or a calculator, we find that the probability corresponding to z = 1.3333 is approximately 0.9082.
Therefore, the probability that a randomly selected orange from the orchard weighs more than 10 oz. is approximately 0.9082 or 90.82% (rounded to the nearest thousandth).