Answer to Question #159430 in Statistics and Probability for TasRak

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 ‘58’ times randomly and the sample average x^- was 7960ppm. 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

Let X be the random variable that represents the water in the output of the process.

Mean μ = 7960

SD σ = 100

n = 58

Find P(X>7950)

Using Central Limit Theorem

$Z = \frac{X-μ}{σ/\sqrt{n}} \\ = \frac{7950-7960}{100/\sqrt{58}} \\ = \frac{-10}{13.14} \\ = -0.761$

The Normal Distribution table give the area to the left of Z.

P(x<Z) = 0.2233

P(x>Z) = 1 – 0.2233

=0.7767

P(X>7950) = 0.7767