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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

Math   Statistics-And-Probability   
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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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