The weight of oranges growing in an orchard is normally distributed with a mean weight of 5 oz. And a standard deviation of 1 oz. Using the empirical rule, determine what interval would represent weights of the middle 95% of all oranges from this orchard.

The interval would represents weights of the middle 95% of all oranges from this orchard.

95%of confidence interval is determined by μ ± 2σ

(5-2, 5+2)

(3,7)

Step-by-step explanation:

Empirical rule:-

The empirical rule states that for a normal distribution

68% of data falls within the first standard deviation from the mean 95% fall within two standard deviations 99.7% fall within three standard deviations

This empirical rule is called the 68-95-99.7 rule

68% of confidence interval is determined by

μ ± σ

95%of confidence interval is determined by

μ ± 2σ

99.7%of confidence interval is determined by

μ ± 3σ

Given data

Mean of the Population μ = 5 o z

Standard deviation of the Population σ = 1 o z

95%of confidence interval is determined by

μ ± 2σ

5 ± 2 (1)

The interval would represents weights of the middle 95% of all oranges from this orchard.

(5 - 2, 5+2)

(3, 7)