QUESTION IMAGE
Question
a set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. sketch a normal curve for the distribution and find the percent of data within each interval.
- between 6.0 and 6.9
- greater than 6.9
- between 4.2 and 6.0
- less than 4.2
- less than 5.1
- between 4.2 and 5.1
test scores are normally distributed with a mean of 76 and a standard deviation of 10.
- in a group of 230 tests, how many students score above 96.
- in a group of 230 tests, how many students score below 66.
- in a group of 230 tests, how many students score within one standard deviation of the mean?
Step1: Calculate the number of standard deviations
For problem 15:
The value \(x = 96\), mean \(\mu=76\), standard deviation \(\sigma = 10\).
The \(z - score=\frac{x-\mu}{\sigma}=\frac{96 - 76}{10}=2\)
For problem 16:
The value \(x = 66\), mean \(\mu = 76\), standard deviation \(\sigma=10\).
The \(z - score=\frac{x-\mu}{\sigma}=\frac{66 - 76}{10}=- 1\)
For problem 17:
The interval within one standard deviation of the mean is \(\mu-\sigma\) to \(\mu+\sigma\), i.e., \(76 - 10\) to \(76+10\) or \(66\) to \(86\). The percentage of data within one standard deviation of the mean in a normal distribution is \(68\%\)
Step2: Use the properties of the normal distribution
For problem 15:
The percentage of data above \(z = 2\) is \(2.5\%\) (using the empirical rule: \(95\%\) of data is within \(\mu\pm2\sigma\), so \(5\%\) is outside, half of which is above \(z = 2\))
The number of students \(n=230\times0.025 = 5.75\approx6\)
For problem 16:
The percentage of data below \(z=-1\) is \(16\%\) (using the empirical rule: \(68\%\) of data is within \(\mu\pm\sigma\), so \(32\%\) is outside, half of which is below \(z=-1\))
The number of students \(n = 230\times0.16=36.8\approx37\)
For problem 17:
The number of students \(n=230\times0.68 = 156.4\approx156\)
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- Approximately \(6\) students score above \(96\).
- Approximately \(37\) students score below \(66\).
- Approximately \(156\) students score within one standard deviation of the mean.