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 4 mg and a standard deviation of 0.1 mg. (Round your answers to four
decimal places.)
What is the probability that a randomly selected capsule contains less than 3.8 mg of vitamin E?
What is the probability that a randomly selected capsule contains at least 4.5 mg of vitamin E?
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To find the probability that a randomly selected capsule contains less than 3.8 mg of vitamin E, we need to calculate the z-score and then use the standard normal distribution table.
z-score = (x - μ) / σ
z-score = (3.8 - 4) / 0.1 = -2
Using the standard normal distribution table, we can find the cumulative probability associated with a z-score of -2. The cumulative probability is the probability that a randomly selected capsule contains less than 3.8 mg of vitamin E.
From the table, we find that the cumulative probability for a z-score of -2 is approximately 0.0228.
Therefore, the probability that a randomly selected capsule contains less than 3.8 mg of vitamin E is approximately 0.0228.
To find the probability that a randomly selected capsule contains at least 4.5 mg of vitamin E, we can calculate the z-score and use the standard normal distribution table.
z-score = (x - μ) / σ
z-score = (4.5 - 4) / 0.1 = 5
Since the z-score of 5 is beyond the values provided in the standard normal distribution table, we can approximate the probability as 1, indicating that it is highly unlikely to find a capsule containing at least 4.5 mg of vitamin E.
Therefore, the probability that a randomly selected capsule contains at least 4.5 mg of vitamin E is approximately 1.
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