show that in the binomial distribution (q+p)^n the ratio of the mean deviation about about the mean to standard deviation is 2/(npq)^0.5*n-1 cµ-1p^µ-1 q^n-µ where µ is the greatest integer contained in np+1 . find the limits of the above ratio as n tends to infinity.
To prove the given result, we need to find the mean and standard deviation of the binomial distribution, and then calculate their ratio.
Let's start by finding the mean deviation about the mean for a binomial distribution. The mean deviation about the mean (MDM) is the average difference between each value in the distribution and the mean.
The mean (μ) of a binomial distribution (q + p)^n is given by np, where n is the number of trials and p is the probability of success in a single trial.
To calculate the mean deviation about the mean, we need to find the difference between each value (k) in the distribution and the mean (np), take the absolute value, and sum them up. Then we divide by the total number of values (n+1). The formula for MDM is as follows:
MDM = (|0 - np| + |1 - np| + |2 - np| + ... + |n - np|) / (n+1)
Now, let's find the standard deviation (σ) of the binomial distribution. The formula for the standard deviation of a binomial distribution is √(npq), where q = 1 - p.
Next, we need to calculate μ, the greatest integer contained in (np + 1).
Finally, we substitute these values into the given ratio:
2/(npq)^0.5 * (n - 1) * c(μ - 1)p^(μ - 1)q^(n - μ)
To find the limits of the above ratio as n tends to infinity, we take the limit of each term separately and then combine them. We subtract 1 from n in the numerator and take the limit as n approaches infinity. We also consider the terms involving p and q as constant since they don't depend on n.
As n approaches infinity, (npq)^0.5 becomes √∞ = ∞.
Thus, the overall limit of the ratio as n tends to infinity is 0.
In conclusion, the limits of the ratio 2/(npq)^0.5 * (n - 1) * c(μ - 1)p^(μ - 1)q^(n - μ) as n tends to infinity is 0.