Answer to Question #241442 in Statistics and Probability for Reyad

Question

The chemical of COVID 19 vaccine is non-toxic to humans. Government

regulations dictate that for any production process involving COVID 19 vaccine,

the water in the output of the process must not exceed 7950 parts per million (ppm)

of COVID 19 vaccine. For a particular process of concern, the water sample was

collected by a manufacturer ‘55’ times randomly and the sample average was 7960

ppm. It is known from historical data that the standard deviation σ is 100 ppm.

What is the probability that the sample average in this experiment would exceed the

government limit if the population mean is equal to the limit? Use the Central Limit

Theorem.

Answer 1

$\mu = 7950 \\ n=55 \\ \bar{x} = 7960 \\ \sigma = 100$

Since sample size is large, according to central limit theorem, we can use normal distribution for sampling distribution of sample means. Therefore:

$P(\bar{x}>7950) = P(\frac{\bar{x} -\mu}{\sigma / \sqrt{n}} > \frac{7960-7950}{100 / \sqrt{55}}) \\ = P(Z> 0.7418) \\ = 1 -P(Z< 0.7418) \\ = 1 -0.7709 \\ = 0.2291$