Question

A drug made by a pharmaceutical company comes in tablet form. Each tablet is branded as containing 95 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 95 mg and standard deviation 2.012 mg.

a)Calculate the percentage of tablets that will contain less than 94 mg of the active chemical. Give your answer as a percentage to 2 decimal places.

Percentage = %

b)Suppose samples of 14 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 94 mg of the active chemical. Give your answer as a percentage to 2 decimal places.

Percentage = %

Answer 1

Solution :

Given that ,

mean = = 95

standard deviation = = 2.012

P(x < 94) = P((x - ) / < (94 - 95) / 2.012) = P(z < -0.4970)

Using standard normal table,

P(x < 94) = 0.3096

Answer = 30.96%

(b)

n = 14

= 95 and

= / n = 2.012 / 14

P( < 94) = P(( - ) / < (94 - 95) / 2.012 / 14 ) = P(z < -1.8597)

Using standard normal table,

P( < 94) = 0.0315

Answer = 3.15%