The distribution of vitamin C amount in vitamin drops produced by a given factory is approximately Normal, with a mean of 60.0 mg and a standard deviation of 0.5 mg.
What is the approximate proportion of vitamin drops containing more than 60.75 mg of vitamin C?

Question

Quizplus · Accepted Answer

To find the proportion of vitamin drops containing more than 60.75 mg of vitamin C, we calculate the z-score for 60.75 mg. The z-score is given by $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. Substituting the given values, we get $z = \frac{(60.75 - 60.0)}{0.5} = 1.5$. Looking up the z-score of 1.5 in a standard normal distribution table or using a calculator, we find that the area to the left of z = 1.5 is approximately 0.9332, which means that about 93.32% of the distribution is below 60.75 mg. Therefore, the proportion above 60.75 mg is 1 - 0.9332 = 0.0668, or approximately 7%.

The Distribution of Vitamin C Amount in Vitamin Drops Produced