Question

The manufacturer of Twitchy Energy Drink is doing a quality control to make sure that the caffeine content (in milligrams, mg) of the drink is meeting specifications. Because of slight natural variation in the bottlin variable with a mean of 74 mg and a standard deviation of 0.3 mg. The inspector takes a sample of 5 bottles from the production line g process, the amount of caffeine in each bottle is actually a random and finds that the average caffeine content in the sample, Y, is 73.8 mg The distribution of caffeine content has historically been reasonably symmetric not following an exact normal distribution. Thus,the central limit theorem will till apply, even though the sample size is smal. Since the product is advertised to provide an energy boost, the company's main concern is having too little caffeine in its product rather than that in any random sample of 5 bottles, the sample mean caffeine content would be 73.8 mg or less? Report your answer to four decimal places Hint: this is a sampling distribution problem.

Answer 1

Let us suppose that X denotes caffeine content

So, X is normally distributed with mean 74 mg and standard deviation 0.3 mg

Now, using the Central Limit theorem,

$\bar{X}$ is normally distributed with mean 74 mg and

$\dpi{100} \small standard\ deviation\ is\ 0.3/\sqrt{5}$

We need to find the Probability that the sample mean caffeine content would be 73.8 mg or less

$\dpi{100} \small P(\bar{X}<73.8) = P\left ( \frac{\bar{X}-Mean}{SD/\sqrt{n}}<\frac{73.8-Mean}{SD/\sqrt{n}} \right )$

$\dpi{100} \small = P\left ( Z<\frac{73.8-74}{0.3/\sqrt{5}} \right )$

$\dpi{100} \small = P\left ( Z<-1.4907 \right )$

$\dpi{100} \small =0.0680$