In a certain distribution of numbers, the mean is 50 with a standard deviation of 5 . At least what fraction of the numbers are between 5 and 95 ? At least $\square$ of all the numbers must lie between 5 and 95 . (Type an integer or a simplified fraction.)

Step 1 ：The question is asking for the fraction of numbers that lie between 5 and 95 in a distribution with a mean of 50 and a standard deviation of 5. This is a question about the properties of a normal distribution.

Step 2 ：According to the empirical rule (or 68-95-99.7 rule), for a normal distribution, about 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and about 99.7% falls within three standard deviations.

Step 3 ：In this case, 5 is 9 standard deviations below the mean and 95 is 9 standard deviations above the mean. Therefore, we can say that almost all (or very close to 100%) of the data falls within this range.

Step 4 ：However, the question asks for the minimum fraction of numbers that must lie within this range. This is a question about Chebyshev's inequality, which provides a lower bound for the amount of data within k standard deviations of the mean for any distribution.

Step 5 ：According to Chebyshev's inequality, at least \(1 - \frac{1}{k^2}\) of the data must be within k standard deviations of the mean.

Step 6 ：In this case, k = 9, so at least \(1 - \frac{1}{81} = \frac{80}{81}\) of the data must be within this range.

Step 7 ：Final Answer: \(\boxed{\frac{80}{81}}\)