Coal is carried from a mine in West Virginia to a power plant in Pittsburgh in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed with an average of 75 tons and a standard deviation of 0.8 ton.

What is the chance 20 cars chosen at random will have an average load of less than 74.7 tons of coal?

1.Enter the average of the box:


2.Enter the SD of the box:

3.Enter the number of draws from the box:

4.Calculate the expected value for the average (EVave):

5.Calculate the standard error for the average (SEave):Round to three decimal places.

6.Calculate the chance 20 cars chosen at random will have an average load of less than 74.7 tons of coal. Round your answer to one decimal place.

You can answer 1-3 just by citing the data.

EVave = sample mean

SEave = SEm = SD/√n

Z = (score-mean)/SEm

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

To calculate the chance that 20 cars chosen at random will have an average load of less than 74.7 tons of coal, we need to follow these steps:

1. Enter the average (mean) of the box: Since the average load for each car is given as 75 tons, we enter 75.

2. Enter the standard deviation (SD) of the box: The standard deviation is given as 0.8 ton, so we enter 0.8.

3. Enter the number of draws from the box: The number of cars chosen at random is given as 20, so we enter 20.

4. Calculate the expected value for the average (EVave): The expected value for the average is the same as the population mean, which is 75.

5. Calculate the standard error for the average (SEave): The standard error for the average is calculated by dividing the standard deviation by the square root of the sample size. In this case, the formula is SEave = SD / sqrt(n) = 0.8 / sqrt(20) ≈ 0.179.

6. Calculate the chance 20 cars chosen at random will have an average load of less than 74.7 tons of coal: To find this probability, we need to calculate the z-score. The z-score is given by the formula (X - EVave) / SEave, where X is the value we are interested in (74.7 tons of coal in this case). Plugging the values into the formula, we get (74.7 - 75) / 0.179 ≈ -0.501.

Using a standard normal distribution table (or a calculator), we can find the area to the left of the z-score -0.501, which represents the probability that the average load of the 20 cars will be less than 74.7 tons of coal. This value is approximately 0.3085, or 30.85%.

So, the chance that 20 cars chosen at random will have an average load of less than 74.7 tons of coal is approximately 30.85%.