Question

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?

Answer 1

To estimate the number of oranges weighing less than 5 oz from a batch of 1500, 50% of the oranges are expected to be below the mean weight of 5 oz, given the normal distribution. Therefore, approximately 750 oranges are expected to weigh less than 5 oz.

Step-by-step explanation:

To determine the number of oranges expected to weigh less than 5 oz in a batch of 1500, we need to understand the properties of the normal distribution. A normal distribution is symmetric around the mean, so exactly half of the distribution lies below the mean and half above. With a mean weight of 5 oz, we would expect that 50% of the oranges weigh less than 5 oz.

In this case, since we have 1500 oranges and half are expected to weigh less than the mean, we can calculate the number expected to weigh less than 5 oz by taking 50% of 1500. Therefore, the calculation would be as follows:

• Number of oranges expected to weigh less than 5 oz = 50% of 1500
• 0.50 * 1500 = 750

To the nearest whole number, we expect 750 oranges to weigh less than 5 oz.

Answer 2

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.