Question

testing a certain pharmaceutical product yielded 5.0 mg of impurity with standard deviation of 3.1 mg....

testing a certain pharmaceutical product yielded 5.0 mg of impurity with standard deviation of 3.1 mg. If 100 samples of pharmaceutical product are prepared, What is the probability that the sample average amount of impurity is between 4.5 mg and 4.8 mg

Homework Answers

Answer #1

Solution :

= / n = 3.1 / 100 = 0.31

= P[(4.5 - 5) / 0.31 < ( - ) / < (4.8 - 5) / 0.31)]

= P(-1.61 < Z < -0.65)

= P(Z < -0.65) - P(Z < -1.61)

= 0.2578 - 0.0537

= 0.2041

Probability = 0.2041

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