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
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?
1 answer