If a group of data has a mean of 19 and a standard deviation of 77.8, what is the interval that should contain at least 88.9% of the data?
Assuming a normal distribution.
If the interval contains 88.9% of the data, we can look up the normal distribution table:
http://cazelais.disted.camosun.bc.ca/264/normal_table.pdf
we find that the half-interval that contains 44.45% of the data is at 1.6σ from the mean.
We extend the interval to both sides of the mean to obtain 88.9% coverage.
Thus, the required interval is
[μ-1.6σ , μ+1.6σ]
where
μ = mean = 19
σ = standard deviation = 77.8
To find the interval that should contain at least 88.9% of the data, we can use the empirical rule, also known as the 68-95-99.7 rule. This rule states that for a normal distribution:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Since we know the standard deviation, we can use this rule to determine the interval.
First, let's find the interval that contains at least 88.9% of the data.
Step 1: Calculate the number of standard deviations needed to capture at least 88.9% of the data.
To do this, we subtract the desired percentage from 100% and divide by 2. In this case:
(100% - 88.9%) / 2 = 5.55%
Step 2: Find the z-score that corresponds to this percentage.
We can use a standard normal distribution table or calculator to find the z-score. In this case, a z-score of 1.65 corresponds to a percentage of 5.55%.
Step 3: Calculate the margin of error.
The margin of error is the z-score multiplied by the standard deviation.
Margin of error = z-score * standard deviation = 1.65 * 77.8
Step 4: Calculate the lower and upper bounds of the interval.
Lower bound = mean - margin of error
Upper bound = mean + margin of error
Therefore, the interval that should contain at least 88.9% of the data is:
(19 - 1.65 * 77.8, 19 + 1.65 * 77.8)