Use the empirical rule to estimate the proportion of values in a specified interval or to find the value that corresponds to a given percentile in a Normal distribution.
68-95-99.7 rule. 1sd, 2sd, or 3sd away from mean, respectively.
The empirical rule, also known as the 68-95-99.7 rule, is a statistical rule of thumb that applies to a normal distribution. According to the rule:
1. Approximately 68% of values in a normal distribution fall within one standard deviation of the mean.
2. Approximately 95% of values in a normal distribution fall within two standard deviations of the mean.
3. Approximately 99.7% of values in a normal distribution fall within three standard deviations of the mean.
Using the empirical rule to estimate the proportion of values in a specified interval involves calculating the z-scores for the endpoints of the interval and then looking up the corresponding proportion from a standard normal distribution table. For example, if we wanted to estimate the proportion of values in the interval (65, 80) for a normal distribution with mean 75 and standard deviation 5, we would first calculate the z-scores for 65 and 80:
z_1 = (65 – 75) / 5 = -2
z_2 = (80 – 75) / 5 = 1
Next, we would look up the proportion of values between z = -2 and z = 1 in a standard normal distribution table, which is approximately 0.818. Therefore, we can estimate that approximately 81.8% of values in the distribution fall between 65 and 80.
To find the value that corresponds to a given percentile in a Normal distribution using the empirical rule, we would use the z-score that corresponds to that percentile and then convert it to the corresponding raw score using the formula x = (z * standard deviation) + mean. For example, to find the value that corresponds to the 90th percentile for a normal distribution with mean 100 and standard deviation 10, we would first find the z-score that corresponds to the 90th percentile, which is approximately 1.28. Then, we would use the formula:
x = (1.28 * 10) + 100 = 112.8
Therefore, we can estimate that the value that corresponds to the 90th percentile in this distribution is approximately 112.8.
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