Multiple choice

An electrical supplier distributor has found the daily demand for fluorescent light bulbs to be normally distributed with a mean of 432 and standard deviation of 86. Find the probability of the demand exceeding 518 bulbs on a particular day.

  1. 0.1587

  2. 0.3413

  3. 0.7587

  4. 0.8413
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A Correct answer
Explanation

Z-score = (X - mean) / SD = (518 - 432) / 86 = 86 / 86 = 1. The probability of Z > 1 is 1 - P(Z < 1) = 1 - 0.8413 = 0.1587.

AI explanation

To find the probability, first calculate the z-score using the formula z equals the observation minus the mean divided by the standard deviation. For a demand of 518 bulbs, z equals 518 minus 432 divided by 86, which equals 1. Using the standard normal distribution table, the probability that z is less than 1 is 0.8413. The probability of the demand exceeding 518 bulbs is 1 minus 0.8413, which equals 0.1587.

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