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.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet