The employees of the company XYZ walk, on average, $26$ km per day, with a standard deviation of $8$ km.
- Assuming a normal distribution for the number of km walked, find the percentage of employees that walk between $20$ and $36$ km per day.
- What can be inferred in absence of information about the distribution?
My solution:
- Since we know that $\mu = 28$ km and $\sigma = 8$ km, we are interested in computing $$ P(20 \leq X \leq 36) $$ Where $X \sim \mathcal{N}(\mu, \sigma)$. If we let $$ Y = \frac{X - \mu}{\sigma} $$ the problem reduces to computing the following probability, where $Y \sim \mathcal{N}(0,1)$: $$ P(-1 \leq Y \leq 1) = $$ $$ P(Y \leq 1) - P(Y \leq -1)= $$ $$ 0.84134 - (1 - 0.84134)= 0.68268 $$ That is the final result.
- In the absence of information about the distribution, we are able to make conclusions using Chebyshev's inequality, i.e.
$$ P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2} $$
Which is useful for finding, for a given $k$, an interval $$ (\mu - k \sigma, \mu + k \sigma) $$ in which the probability of finding a realization of the random variable $X$ doesn't exceed $\frac{1}{k^2}$.
As an example, by setting $k = 1.2103$, we know that the probability of observing values of $X$ outside the interval $$ (28 - 1.2103 \times 8, 28 + 1.2103 \times 8) = $$ $$ (18.3176, 37.6824) $$ Is equal to $0.68267$ (rounded to the fifth decimal digit.)
Are my conclusions and computations correct? I hope that somebody is willing to take some time to help :)
You are interested in computing
$$\mathbb{P}(20 \leq X \leq 36)= \mathbb{P}\left(\frac{20-28}{8} \leq \frac{X-28}{8}\leq \frac{36-28}{8}\right) = \mathbb{P}(-1 \le Y \le1)$$
so (a) is correct (I get the same end value as you do).
(b) is an open question and I'm not sure what your instructor wants to see here. Maybe you can use the central limit theorem or something like that to approximate a normal variable, but what you did seems ok.