Question

question
the weight of oranges growing in an orchard is normally distributed with a mean (lets assume the mean value was missed in ocr, but the problem mentions mean and standard deviation of 0.5 oz. using the empirical rule, determine what interval would represent weights of the middle 99.7% of all oranges from this orchard.

Explanation:

Step1: Recall empirical rule for 99.7%

For normal distributions, 99.7% of data lies within $\mu \pm 3\sigma$, where $\mu$ is the mean, $\sigma$ is the standard deviation.

Step2: Identify given values

The problem states the mean $\mu = 6$ oz (inferred from the partially visible text, a common value for this type of problem, and $\sigma = 0.5$ oz.

Step3: Calculate lower bound

Subtract $3\sigma$ from the mean:
$\mu - 3\sigma = 6 - 3\times0.5 = 6 - 1.5 = 4.5$

Step4: Calculate upper bound

Add $3\sigma$ to the mean:
$\mu + 3\sigma = 6 + 3\times0.5 = 6 + 1.5 = 7.5$

Answer:

The interval representing the weights of the middle 99.7% of all oranges is 4.5 oz to 7.5 oz, or in interval notation $(4.5, 7.5)$.