The weight of oranges growing in an orchard is normally distributed with a mean weight of 6 oz. and a standard deviation of 0.5 oz. Using the empirical rule, determine what interval would represent weights of the middle 95% of all oranges from this orchard.

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.

x = 23.78

step-by-step explanation:

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: