Question

scores on a standardized intelligence test are normally distributed with a mean of 100 and a standard deviation of 18. use the 68 - 95 - 99.7 rule to find the percentage of people taking the test who score below 64

the percentage of people taking the test who score below 64 is %
(type an integer or a decimal.)

Explanation:

Step1: Calculate the difference between the mean and the given score

The mean $\mu = 100$, the score $x = 64$. The difference is $100 - 64=36$.

Step2: Find how many standard deviations the score is from the mean

The standard deviation $\sigma = 18$. The number of standard deviations $n=\frac{36}{18} = 2$.

Step3: Apply the 68 - 95 - 99.7 Rule

The 68 - 95 - 99.7 Rule states that for a normal distribution:

• Approximately 68% of the data lies within $1$ standard deviation of the mean ($\mu\pm\sigma$).
• Approximately 95% of the data lies within $2$ standard deviations of the mean ($\mu\pm2\sigma$).
• Approximately 99.7% of the data lies within $3$ standard deviations of the mean ($\mu\pm3\sigma$).

The percentage of data within $2$ standard deviations of the mean is 95%. So the percentage of data outside of $2$ standard deviations is $100 - 95=5\%$. Since the normal distribution is symmetric, the percentage of data less than $\mu - 2\sigma$ is $\frac{100 - 95}{2}=2.5\%$.

Answer:

$2.5$