Binomial probability - find the probability that an observation will fall into each of these intervals

[Herondaleheir]

Let X be a binomial random variable with n = 20 and p = 0.1

Calculate the intervals μ ± σ, μ ± 2σ, and μ ± 3σ. Find the probability that an observation will fall into each of these intervals.

So I know the mean is 2 and the standard deviation is 1.342. I can calculate the actual intervals for each standard deviation, but I don't get how to calculate the actual probability for each of them and would appreciate some help.

Like for μ ± σ I know the interval is (0.658, 3.342) but I don't get how to turn that into a probability?

The textbook states the answers are 0.7455, 0.9569, 0.9977

 

[Romsek]

$\displaystyle \text{$X$ can only take integral values from 0 to 20}$

$\displaystyle \text{In the interval $(0.658, 3.342)$ there are 3 valid values of $X,~X=1,2,3$}$

$\displaystyle \text{Sum the probabilities of each of those values using $Pr[k] = \dbinom{20}{k}(0.1)^k(0.9)^{20-k}$}$

$\displaystyle \text{Repeat this for each interval}$

 

[JeffM]

[MATH]2 - 1.342 = 0.658 \text { and } 2 + 1.342 = 3.342.[/MATH]
So what is the probability that the number of sucesses s is

[MATH]0.658 \le s \le 3.342 \implies 1 \le s \le 3 \ \because \ s \in \mathbb Z_{\ge 0}.[/MATH]
[MATH]\dbinom{20}{1} * 0.1 * 0.9^{19} + \dbinom{20}{2} * 0.1^2 * 0.9^{18} + \dbinom{20}{3} * 0.1^3 * 0.9^{17} \approx 0.74547.[/MATH]
Same idea as romsek. Things get a bit more interesting for the other intervals.