A desktop PC uses an average of 120 watts of electricity per hour, based on 4 hours of use per day. Assume the variable is approximately normal and the standard deviation is 6 watts. If 500 PCs are selected at random, approximately how many will use more than 130 watts?

A test prep company is targeting students that scored in the bottom 5% on a certain test. If the mean score on the 100-point test was 68 with a standard deviation of 12, what would be the highest score in the lowest 5%? Assume that test scores are approximately normal.

Approximately 24 out of 500 PCs are expected to have power consumptions exceeding 130 watts. The highest score in the bottom 5% of a normally distributed test score with a mean of 68 and a standard deviation of 12 is approximately 48.

Explanation:

The first part of your question deals with finding how many PCs out of 500 are expected to use more than 130 watts. This is a problem of statistics and probability, specifically, it's about finding the proportion of a normal distribution that lies above a certain value. Let's start by finding the z-score for 130 watts (z = (X - μ) / σ), which is approximately 1.67. Next, we calculate the proportion of PCs expected to use more than 130 watts by subtracting the value of the standard normal cumulative distribution at 1.67 from 1. This gives us approximately 0.0475. Multiplying 0.0475 by 500 gives us about 24 PCs expected to have power consumption more than 130 watts.

The second part of your question refers to finding the maximum score in the lowest 5% of a normally distributed test score. We need to find the z-score that corresponds to the 5th percentile and then use z = (X - μ) / σ to find the corresponding test score. The z-score for the 5th percentile is approximately -1.645. Solving the equation X = μ + z * σ for X gives us a score of around 48. Thus, the highest score in the bottom 5% of test scores is approximately 48.

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