Answer:
The interval that would represent weights of the middle 95% of all oranges from this orchard is from 5 oz to 7 oz.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 6
Standard deviation = 0.5
Middle 95% of weights:
By the Empirical Rule, within 2 standard deviations of the mean.
6 - 2*0.5 = 5
6 + 2*0.5 = 7
The interval that would represent weights of the middle 95% of all oranges from this orchard is from 5 oz to 7 oz.