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the mean of a certain distribution is 100 with a standard deviation of …

Question

the mean of a certain distribution is 100 with a standard deviation of 20.
according to the 68-95-99.7 rule, what percent of the data would fall between 80 and 120?

Explanation:

Step1: Identify the mean and standard deviation

The mean (\(\mu\)) is 100, and the standard deviation (\(\sigma\)) is 20.

Step2: Calculate the number of standard deviations from the mean for 80 and 120

For 80: \( \frac{80 - \mu}{\sigma} = \frac{80 - 100}{20} = -1 \)
For 120: \( \frac{120 - \mu}{\sigma} = \frac{120 - 100}{20} = 1 \)

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 falls within \( \mu\pm\sigma \), 95% within \( \mu\pm2\sigma \), and 99.7% within \( \mu\pm3\sigma \). Since 80 is \( \mu - \sigma \) and 120 is \( \mu+\sigma \), the percentage of data between 80 and 120 is approximately 68%.

Answer:

68%