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Question
the scores on a psychology exam were normally distributed with a mean of 60 and a standard deviation of 4. a failing grade on the exam was anything 2 or more standard deviations below the mean. what was the cutoff for a failing score? approximately what percentage of the students failed?
the cutoff for a failing score was
(simplify your answer.)
Step1: Calculate the cutoff score
The formula for a value \(x\) in a normal distribution is \(x=\mu - z\sigma\), where \(\mu\) is the mean, \(z\) is the number of standard deviations, and \(\sigma\) is the standard deviation.
Given \(\mu = 60\), \(z = 2\), \(\sigma=4\).
Substitute the values into the formula: \(x=60-2\times4\).
Step2: Determine the percentage of students who failed (using the empirical rule for normal distributions)
The empirical rule (68 - 95 - 99.7 rule) states that for a normal distribution:
- Approximately 95% of the data lies within \(z=\pm2\) standard deviations of the mean.
- The total area under the normal - distribution curve is 1 (or 100%).
The area outside of \(z = \pm2\) is \(100\%-95\% = 5\%\).
Since the distribution is symmetric, the area below \(z=- 2\) (the failing area) is \(\frac{100\% - 95\%}{2}=2.5\%\)
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The cutoff for a failing score was \(52\). Approximately \(2.5\%\) of the students failed.