Question

Monthly fuel consumption at a firm is normally distributed with mean 20,000 liters and standard deviation of 2,500 liters. suppose the firm wants to store fuel in a tank. what capacity does this tank need to have to store monthly fuel consumption with 0.90 probability? round your answer to the nearest integer.

Answer 1

The capacity of the tank needed to ensure a 90% probability of covering monthly fuel consumption at the firm is 23,200 liters.

Step-by-step explanation:

To find the capacity of the fuel tank necessary to ensure that 90% of the monthly fuel consumption at the firm is covered, we use the concept of a normal distribution and the Z-score associated with a 90% probability. Given a mean (μ) of 20,000 liters and a standard deviation (σ) of 2,500 liters, we look for the Z-score that corresponds to 90% probability in a Z-table, which is approximately 1.28. Next, we calculate the capacity using the formula:

X = μ + Z * σ

where X is the desired capacity. Plugging in the values, we get:

X = 20,000 + 1.28 * 2,500

X = 20,000 + 3,200

X = 23,200

Therefore, the capacity that the tank needs to have to store at least 90% of the monthly fuel consumption is 23,200 liters, when rounded to the nearest integer.

Answer 2

The tank needs to have a capacity of 23,204 liters.

Monthly fuel consumption at the firm is normally distributed with mean 20,000 liters and standard deviation of 2,500 liters. The firm wants to store fuel in a tank that can store monthly fuel consumption with 0.90 probability.

To do this, we need to find the tank capacity that corresponds to the 90th percentile of the normal distribution.

The 90th percentile of the standard normal distribution is 1.282. To find the 90th percentile of the normal distribution with mean 20,000 and standard deviation 2,500, we can use the following formula:

z = (x - mean) / standard deviation

where:

z is the z-score

x is the value we want to find (the 90th percentile)

mean is the mean of the normal distribution (20,000)

standard deviation is the standard deviation of the normal distribution (2,500)

Solving for x, we get:

x = mean + z * standard deviation

x = 20,000 + 1.282 * 2,500

x = 23,204

Therefore, the tank needs to have a capacity of at least 23,204 liters to store monthly fuel consumption with 0.90 probability.