A cola dispensing machine is set to dispense a mean of 2.02 liters into a container labeled 2 liters. Actual quantities dispensed vary and the amounts are normally distributed with a standard deviation of 0.015 liter. What is the probability a container will have less than 2 liters?
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A. B. C. D. E.B
z = (x-u)/sigma = 2 - 2.02/0.015 = -1.33 (negative since it is below the mean). The area between - infinity and z = -1.33 is 0.0918.
To solve this problem, we need to use the concept of the standard normal distribution and the Z-score. The Z-score measures the number of standard deviations an observation or data point is from the mean.
Given that the mean amount dispensed is 2.02 liters and the standard deviation is 0.015 liters, we can calculate the Z-score for the desired quantity of 2 liters.
Z-score formula: Z = (X - μ) / σ
Where: Z = Z-score X = Desired quantity μ = Mean σ = Standard deviation
Plugging in the values, we have: Z = (2 - 2.02) / 0.015
Calculating this, we get: Z = (-0.02) / 0.015 Z = -1.33 (rounded to two decimal places)
To find the probability of a container having less than 2 liters, we need to find the area under the standard normal distribution curve to the left of the Z-score. We can refer to a Z-table or use statistical software to find this value.
Looking up the Z-table or using a calculator, we find that the cumulative probability for a Z-score of -1.33 is approximately 0.0918.
Therefore, the probability that a container will have less than 2 liters is approximately 0.0918.
The correct answer is B. 0.0918.