Chebyshev's inequality, how is it applied in this problem?

Question

Data set: Problem:

Work attempted:

There is no way that .21 % is at least 3.48, especially when 29/30 are between the bounds of (3.48, 3.96). It's not clear to me what I'm doing wrong.

Answer 1

Haven't worked through the numbers, but I will describe what I think is the procedure (I use slightly different notation since I treat the data as a population).

You are given a bunch of data $x_1,...,x_n.$ For the probabilistic setup in this question, treat these as realizations of a random variable $X$, where each observation has uniform probability mass of $1/n$.

1. Compute the mean $\mu=\frac{1}{n}\sum_i x_i$ and variance $\sigma^2=\frac{1}{n}\sum_i (x_i-\mu)^2=(\frac{1}{n}\sum_i x_i^2)-\mu^2.$

(It's unclear if the problem intended you to use the sample variance, $\frac{1}{n-1}\sum_i (x_i-\bar x)^2$. Treating the data as the population, I don't see how using sample variance makes much sense. However, since $n$ is somewhat large, I suppose which one you choose won't make too much of a difference for the end result).

2. Count the number of observations in the interval $(\mu-1.5\sigma,\mu+1.5\sigma)$ and call the count $n_1$, i.e.

$$n_1\equiv \#\{x_i: |x_i-\mu|< 1.5\sigma\}.$$

The proportion of data in said interval is then computed as $ p\equiv n_1/n.$

Note that since each observation is modeled as having mass $1/n$, we can equivalently write $$p=P(|X-\mu|<1.5\sigma).$$

3. Chebyshev's ineq. says

$$P(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2}\iff p= P(|X-\mu|< k\sigma)\geq 1-\frac{1}{k^2}$$

So if all works well, you should find $p\geq1-\frac{1}{1.5^2}= 0.\bar 5$.