Question

if the mean of a standardized test with a normal distribution is 54.3 and the standard deviation is 4.6, what is the best approximation of the percent of the scores that fall between 54.3 and 63.5?
34
47.5
95
68

Explanation:

Step1: Calculate the number of standard deviations

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 54.3\) (mean), \(\sigma=4.6\) (standard deviation), and \(x = 63.5\).
Substitute the values: \(z=\frac{63.5 - 54.3}{4.6}=\frac{9.2}{4.6}=2\)

Step2: Use the empirical rule for normal distribution

The empirical rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(z=\pm1\)
• Approximately \(95\%\) of the data lies within \(z=\pm2\)
• Approximately \(99.7\%\) of the data lies within \(z=\pm3\)

We want the percentage between \(z = 0\) (since \(x=\mu = 54.3\) gives \(z = 0\)) and \(z = 2\).
Since the normal distribution is symmetric about the mean (\(z = 0\)), the percentage of data between \(z = 0\) and \(z=2\) is \(\frac{95\%}{2}=47.5\%\)

Answer:

47.5