In a normal distribution, we can use z-scores to determine probabilities. The z-score formula is z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.
For a test score of 90, the z-score would be calculated as follows:
z = (90 - 79) / 5.5 = 11 / 5.5 = 2
Using a z-table, we can determine that the probability of a z-score being greater than 2 is approximately 0.0228.
Therefore, the probability of a randomly selected test score being more than 90 is approximately 2.28%, which is closest to the option 5%.
The mean score on a particular test is 79 and the standard deviation is 5.5What is the probability that a randomly selected test score is more than 90point) 34% 68% 5% 95%
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