The weight of oranges growing in an orchard is normally distributed with a mean weight of 5 oz. and a standard deviation of 0.5 oz. From a batch of 1500 oranges, how many would be expected to weight less than 5 oz., to the nearest whole number?

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.

1/6

Step-by-step explanation:

total = 12

number of butterscotch = 4

number of mint = 6

Since he replaces the mint, these events are independent events.

p(butterscotch, then mint) = p(butterscotch) * p(mint)

p(event) = (number of desired outcomes)/(total number of outcomes)

p(butterscotch) = 4/12 = 1/3

p(mint) = 6/12 = 1/2

p(butterscotch, then mint) = p(butterscotch) * p(mint)

p(butterscotch, then mint) = 1/3 * 1/2

p(butterscotch, then mint) = 1/6

1/6