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.

does this formula look familar?

z= (x-mean)/(standard deviation)

Find the z-score that matches 4%. Then, plug in that for z. Finally, solve for standard deviation.

To find the standard deviation of the normal distribution in this problem, we need to use the concept of the Z-score. The Z-score is the number of standard deviations a data point is away from the mean.

In this case, we know that 4 percent of the cups overflow. This means that 96 percent of the cups do not overflow. Since the normal distribution is symmetric, we can divide the remaining 96 percent equally on both sides of the mean.

To find the Z-score corresponding to the 96th percentile, we can use a Z-table or a calculator. The Z-score corresponding to the 96th percentile is approximately 1.75. This means that the cups overflowing have a Z-score of approximately 1.75.

Now, we can use the Z-score formula to find the standard deviation:

Z = (X - μ) / σ,

where Z is the Z-score, X is the value, μ is the mean, and σ is the standard deviation.

Since the Z-score corresponding to the cups overflowing is 1.75 and the mean is 7 oz, we can rearrange the formula to solve for the standard deviation:

1.75 = (8 - 7) / σ.

Simplifying the equation:

1.75 = 1 / σ.

To isolate σ, we take the reciprocal of 1.75:

σ = 1 / 1.75 = 0.5714.

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