Question

question 5
a given set of scores forms a normal distribution. the mean of the set of scores is 80 and the standard deviation is 6. what percentage of the scores lie between 80 and 86?
47.5
34
95
68

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For \(x = 86\), \(\mu=80\), \(\sigma = 6\), then \(z=\frac{86 - 80}{6}=\frac{6}{6}=1\)

Step2: Use the empirical rule (68 - 95 - 99.7 rule)

In a normal distribution:

• Approximately 68% of the data lies within \(z=- 1\) and \(z = 1\) (i.e., \(\mu\pm\sigma\))
• Approximately 95% of the data lies within \(z=-2\) and \(z = 2\) (i.e., \(\mu\pm2\sigma\))
• Approximately 99.7% of the data lies within \(z=-3\) and \(z = 3\) (i.e., \(\mu\pm3\sigma\))

The normal distribution is symmetric about the mean (\(z = 0\)). The percentage of data between \(z = 0\) and \(z = 1\) is half of the percentage of data between \(z=-1\) and \(z = 1\)

Since the percentage of data between \(z=-1\) and \(z = 1\) is 68%, the percentage of data between \(z = 0\) (mean) and \(z = 1\) is \(\frac{68\%}{2}=34\%\)

Answer:

34