scores on a standardized intelligence test are normally distributed with a mean of 100 and a standard…

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.\nthe percentage of people taking the test who score above 128 is (%). (type an integer or a decimal.)

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.\nthe percentage of people taking the test who score above 128 is (%). (type an integer or a decimal.)

Answer

Explanation:

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., $P(\mu - 2\sigma<X<\mu + 2\sigma)\approx0.95$. The total area under the normal curve is 1. The area outside of $\mu\pm2\sigma$ is $1 - 0.95 = 0.05$. Since the normal distribution is symmetric, the area above $\mu + 2\sigma$ is $\frac{1 - 0.95}{2}=\frac{0.05}{2}=0.025$.

Answer:

2.5