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Question
scores on a standardized intelligence test are normally distributed with a mean of 100 and a standard deviation of 14. use the 68 - 95 - 99.7 rule to find the percentage of people taking the test who score above 128.
the percentage of people taking the test who score above 128 is (%). (type an integer or a decimal.)
Step1: Calculate number of standard - deviations
The mean $\mu = 100$ and standard deviation $\sigma=14$. We want to find how many standard - deviations 128 is from the mean. Let $x = 128$. The number of standard - deviations $z=\frac{x - \mu}{\sigma}=\frac{128 - 100}{14}=\frac{28}{14}=2$.
Step2: Apply the 68 - 95 - 99.7 Rule
The 68 - 95 - 99.7 Rule states that about 95% of the data lies within 2 standard - deviations of the mean, i.e., $P(\mu - 2\sigma
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