A soft drink vending machine is supposed to fill cups with pop. The cups hold exactly 8 oz of liquid. Suppose the actual amount dispensed by the machine follows a normal distribution with mean 7 oz. If it is found that exactly 4 percent of the cups overflow, what is the standard deviation of the normal distribution??

omg, please help.

A vending machine is designed to dispense a mean of 7.6 oz. of coffee into an 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.4 oz. and the amount is normally distributed, find the percent of times that the machine will:

a.) Dispense from 7.4 oz. to 7.7 oz.
b.) Dispense less than 7.0 oz.

To find the standard deviation of the normal distribution, we need to solve the problem with the given information.

In this case, we know that the mean (μ) of the distribution is 7 oz.

Let's assume X as the amount dispensed by the machine, and we want to find the standard deviation (σ) of X. Since the problem states that exactly 4 percent of the cups overflow, we can see that we are looking for the X value that is 4 percent above the cup capacity of 8 oz.

To find this X value, we can use the Z-score formula. The Z-score formula allows us to convert X values from normal distribution to standard normal distribution, which has a mean of 0 and standard deviation of 1. We can then use the Z-score to find the corresponding percentile.

The formula for Z-score is:
Z = (X - μ) / σ

Since we want to find the X value that corresponds to the 96th percentile (100% - 4%) of the distribution, we can look up the Z-score value for that percentile in a standard normal distribution table.

The Z-score value for the 96th percentile is approximately 1.75, which means that the corresponding X value is:
1.75 = (X - 7) / σ

We know that X should be 8 oz, so we can substitute these values into the equation:
1.75 = (8 - 7) / σ
1.75 = 1 / σ

Now we can solve for σ:
1.75σ = 1
σ = 1 / 1.75
σ ≈ 0.5714

Therefore, the standard deviation of the normal distribution is approximately 0.5714 oz.