Question

6. The fill amount in wine bottles is normally distributed, with a mean of 750 ml and a standard deviation of 18 ml. Bottles that have more than 772.5 ml may cause spills upon opening. Normal Distribution Table: a. What is the probability that a bottle may spill upon opening? b. The probability is 90% that the amount of wine contained in a bottle is higher than what fill amount?

Answer 1

Answer 6 a:

Given,

Mean ($\mu$) = 750 ml

Standard Deviation ($\sigma$) = 18 ml

Z score = (X - $\mu$ ) / $\sigma$ = (772.5 - 750) / 18 = 1.25

Probability that bottle may spill upon opening = Probability (Z>1.25) = 0.1826 or 18.26%

Answer 6 b:

Given,

Mean ($\mu$) = 750 ml

Standard Deviation ($\sigma$) = 18 ml

Let X ml be the the fill amount such that the amount of wine in the bottle has a probability of 90%

Z Score = -1.2816 [calculated for probability of (1 - 0.9) = 0.1 or 10%]

Using the equation, Z score = (X - $\mu$ ) / $\sigma$ ,

X = Z score * $\sigma$ + $\mu$ = 726.93 ml.

Therefore, the fill amount such that the amount of wine in the bottle has a probability of 90% is 726.93 ml.