Question

A drug made by a pharmaceutical company comes in tablet form. Each tablet is branded as containing 105 mg of the particular active chemical. However, variation in manufacturing results in the actual amount of the active chemical in each tablet following a normal distribution with mean 105 mg and standard deviation 1.513 mg.

A) Calculate the percentage of tablets that will contain less than 104 mg of the active chemical.
B) Suppose samples of 15 randomly selected tablets are taken and the amount of active chemical measured. Calculate the percentage of samples that will have a sample mean of less than 104 mg of the active chemical.

Answer 1

a) 25.46% of tablets will contain less than 104 mg of the active chemical.

b) 0.52% of samples will have a sample mean of less than 104 mg of the active chemical.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normal distribution with mean 105 mg and standard deviation 1.513 mg.

This means that
`\mu = 105, \sigma = 1.513`

A) Calculate the percentage of tablets that will contain less than 104 mg of the active chemical.

The proportion is the pvalue of Z when X = 104. So

`Z = (X - \mu)/(\sigma)`

`Z = (104 - 105)/(1.513)`

`Z = -0.66`

`Z = -0.66` has a pvalue of 0.2546

0.2545*100 = 25.46

25.46% of tablets will contain less than 104 mg of the active chemical.

B) Suppose samples of 15 randomly selected tablets are taken and the amount of active chemical measured. Calculate the percentage of samples that will have a sample mean of less than 104 mg of the active chemical.

Sample means from samples of 15. So, by the Central Limit Theorem:

`n = 15, s = (1.513)/(√(15)) = 0.39`

The proportion is the pvalue of Z when X = 104. So

`Z = (X - \mu)/(s)`

`Z = (104 - 105)/(0.39)`

`Z = -2.56`

`Z = -2.56` has a pvalue of 0.0052

0.0052*100 = 0.52

0.52% of samples will have a sample mean of less than 104 mg of the active chemical.