200k views
4 votes
Now suppose that bigger cups are ordered and the machine’s mean amount dispensed is set at μ=12. Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?

1 Answer

5 votes

Answer:

σ should be adjusted at 0.5.

Explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean 12.

Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?

13 should be 2 standard deviations above the mean of 12, and 11 should be two standard deviations below the mean.

So 1 should be worth two standard deviations. Then


2\sigma = 1


\sigma = (1)/(2)


\sigma = 0.5

σ should be adjusted at 0.5.

User AMA
by
8.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories