Binomials for getting probability of standard deviation

Question

I have the following problem which I am stuck on the second part.

Suppose that $30\%$ of all students who have to buy a text for a particular course want a new copy whereas the other $70\%$ want a used copy. Consider randomly selecting $25$ purchasers.

a. What are the mean value and standard deviation of the number who want a new copy of the book?

b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value?

For part a I did the following:

$\text{mean} = 25(.30) = 7.5$

$\text{standard deviation} = \sqrt{25 \cdot .30 \cdot .70} = 2.291$

For part b however I am stuck. I saw online a similar problem where someone was doing $P(x <1.9) + P(x > 10.1)$ to get their answer, but I have no clue how they came to this conclusion or how to solve $P(x<1.9) + P(x>10.1)$

Answer 1

(b) is asking the following: What is the chance that the number of people wanting new copies is either:

• 2 standard deviations above the mean: this means above $7.5+2\cdot 2.291=12.082$, so above or equal to 13 people, in symbols this is $P(x>12.082)=P(x\geq 13)=$ $=P(x=13)+P(x=14)+\dots+P(x=24)+P(x=25)$,
• or 2 standard deviations below the mean, so below $7.5-2\cdot 2.291=2.918$, so below or equal to 2 people, in symbols: $P(x< 2.918)=P(x\leq 2) = P(x=1)+P(x=2)$.

All together, the probability asked is: $P(x\leq 2)+P(x\geq 13)= P(x<3)+P(x>12)=$

$P(x=1)+P(x=2)+P(x=13)+P(x=14)+\dots+P(x=24)+P(x=25)$

Now, that's a lot to calculate, it can be a liiittle shortened if you remember that the above sum is equal to:

$1-P(2<x<13)=1-(P(x=3)+\dots +P(x=12))$

But still to me it seems unnecessarily too much to compute in order to learn the binomial distribution... Maybe you have calculators that can compute this kind of sum quickly?