A machine producing vitamin E capsules operates so that the actual amount of vitamin E in each capsule is normally distributed with a mean of 6 mg and a standard deviation of 0.02 mg. (You may need to use a table or technology. Round your answers to four decimal places.) What is the probability that a randomly selected capsule contains less than 5.98 mg of vitamin E? What is the probability that a randomly selected capsule contains at least 6.08 mg of vitamin E?
Added by Timothy J.
Step 1
The z-score is calculated by subtracting the mean from the value and dividing by the standard deviation. For 5.98 mg, the z-score is (5.98 - 6) / 0.02 = -1. For 6.08 mg, the z-score is (6.08 - 6) / 0.02 = 4. Show more…
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