A bottler of drinking water fills plastic bottles with a mean volume of 998 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally distributed. What
Proportion of bottles have volumes greater than 999 mL?
A) 0.8051
B) 1.0000
C) 0.6293
D) 0.4443

Question

Quizplus · Accepted Answer

To find the proportion of bottles with volumes greater than 999 mL, we use the Z-score formula: Z = (X - μ) / σ, where X is the value in question (999 mL), μ is the mean (998 mL), and σ is the standard deviation (7 mL). Plugging in the values gives Z = (999 - 998) / 7 = 1 / 7 ≈ 0.143. Looking up this Z-score in a standard normal distribution table or using a calculator, we find that the area to the left of this Z-score is approximately 0.5567, meaning the area to the right (which represents the proportion of bottles with volumes greater than 999 mL) is 1 - 0.5567 = 0.4433, which rounds to 0.4443.

A Bottler of Drinking Water Fills Plastic Bottles with a Mean