Daily output of Marathon's Grayville, LA, Refinery is normally Distributed with a mean of 232,000 barrels of crude oil per day with a standard deviation of 7,000 barrels.
a. what is the probability of producing at least 232,000 barrels?
b. between 232,000 and 239,000 barrels?
c. Less than 239,000 barrels?
d. Less than 245,000 barrels?
e. More than 225,000 barrels?
To find the probabilities in this scenario, we can use the Standard Normal Distribution with the given mean and standard deviation. We will calculate the z-scores for the given values and then use a table or calculator to find the corresponding probabilities.
a. Probability of producing at least 232,000 barrels:
To find this probability, we need to find the area to the right of 232,000 on the Standard Normal Distribution curve.
Step 1: Calculate the z-score:
z = (x - mean) / standard deviation
z = (232,000 - 232,000) / 7,000
z = 0
Step 2: Find the probability using the z-score:
Since the z-score is 0, we need to find the area to the right of z = 0. According to the Standard Normal Distribution table, this probability is 0.5.
Therefore, the probability of producing at least 232,000 barrels is 0.5 or 50%.
b. Probability of producing between 232,000 and 239,000 barrels:
To find this probability, we need to find the area between 232,000 and 239,000 on the Standard Normal Distribution curve.
Step 1: Calculate the z-scores:
For 232,000:
z1 = (232,000 - 232,000) / 7,000
z1 = 0
For 239,000:
z2 = (239,000 - 232,000) / 7,000
z2 = 1
Step 2: Find the probability using the z-scores:
We need to find the area between z1 = 0 and z2 = 1. Using the Standard Normal Distribution table or a calculator, we find that the probability is approximately 0.3413.
Therefore, the probability of producing between 232,000 and 239,000 barrels is approximately 34.13%.
c. Probability of producing less than 239,000 barrels:
To find this probability, we need to find the area to the left of 239,000 on the Standard Normal Distribution curve.
Step 1: Calculate the z-score:
z = (239,000 - 232,000) / 7,000
z ≈ 1
Step 2: Find the probability using the z-score:
We need to find the area to the left of z ≈ 1. According to the Standard Normal Distribution table, this probability is approximately 0.8413.
Therefore, the probability of producing less than 239,000 barrels is approximately 84.13%.
d. Probability of producing less than 245,000 barrels:
To find this probability, we need to find the area to the left of 245,000 on the Standard Normal Distribution curve.
Step 1: Calculate the z-score:
z = (245,000 - 232,000) / 7,000
z ≈ 1.86
Step 2: Find the probability using the z-score:
We need to find the area to the left of z ≈ 1.86. According to the Standard Normal Distribution table, this probability is approximately 0.9693.
Therefore, the probability of producing less than 245,000 barrels is approximately 96.93%.
e. Probability of producing more than 225,000 barrels:
To find this probability, we need to find the area to the right of 225,000 on the Standard Normal Distribution curve.
Step 1: Calculate the z-score:
z = (225,000 - 232,000) / 7,000
z ≈ -1
Step 2: Find the probability using the z-score:
We need to find the area to the right of z ≈ -1. According to the Standard Normal Distribution table, this probability is approximately 0.1587.
Therefore, the probability of producing more than 225,000 barrels is approximately 15.87%.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.