Probability and Statistics Binomials distribution I am having trouble understanding a problem that I was given today, I was supplied with the answers but no explanation so I am trying to figure out how to get these answers. So heres the queston:
A manufacturer knows that on average 20% of the electric toasters produced require repairs within 1 year after they are sold. When 20 toasters are randomly selected, find appropriate numbers of x and y such that. (Do not use normal distribution to solve)
a) the probability that at least x of them will require repairs is less than 0.5 
Apparently the answer is 4 but I am not sure why I was told it was because it is the mean but I do not get how the question is even asking for that.
b)the probability that at least y of them will not require repairs is greater than 0.8 
For this one I was supplied an answer using normal distribution even though we are told not to use that...
Things that are known:
P=0.2, q=0.8, n=20, x=# of toasters that require repair.
Any help with trying to understand how to answer these questions would be greatly appreciated!
 A: You need to use the cumulative binomial distribution.  It should be easy to locate a table somewhere.  Such tables typically go up to $n=20$ or more and give the distribution for various values of $p$ between $0$ and $1,$ typically spaced by $0.05$ or $0.1$.  For a given $n$ and $p$ the table will give the probability that $k$ or fewer successes occur.  The complement of this probability is the probability that $k+1$ or more successes occur.
The suggestion to use the fact that the mean is $4$ to answer the first question apparently is relying on the mean and the median being close to each other in the binomial distribution.  Wikipedia lists some results relating the median and the mean.  To me, this information seems tricky to apply in general.  In fact, I don't think the person who gave you that answer got it right.  Cumulative binomial tables are probably a better tool for small $n,$ and the normal approximation is a better tool for large $n.$
Added:  If you know how to compute the cumulative distribution function of the binomial distribution by hand or by some other method, then you can use that method on this problem too.  To keep the numbers simple, suppose that the sample has $5$ toasters.  Let $K$ be the number of toasters requiring repair, so that $M=5-K$ is the number not requiring repair. Then
$$
\begin{array}{lll}
\Pr[K \le 0]=\Pr[M\ge5]=0.32768 & \phantom{AAAA} & \Pr[K\ge1]=0.67232\\
\Pr[K \le 1]=\Pr[M\ge4]=0.73728 & \phantom{AAAA} & \Pr[K\ge2]=0.26272\\
\Pr[K \le 2]=\Pr[M\ge3]=0.94208 & \phantom{AAAA} & \Pr[K\ge3]=0.05792\\
\Pr[K \le 3]=\Pr[M\ge2]=0.99328 & \phantom{AAAA} & \Pr[K\ge4]=0.00672\\
\end{array}
$$
Question (a) can be answered by looking at the second column of numbers.  Since the probability that one or more toasters require repair is greater than $0.5$ but the probability that two or more require repair is less than $0.5,$ we conclude that $x=2.$  
Question (b) can be answered by looking at the first column of numbers.  Since the probability that at least four do not require repair is less than $0.8$ but the probability that at least three do not require repair is greater than $0.8,$ we conclude that $y=3.$
A table like the one here can save you the trouble of computing the first column of numbers yourself.  (The second column will have to be computed, but it's just the complement of the first column.)
A: Suppose that the number of toasters that need to be repaired after 1 year is represented by the random variable $X$, and also suppose that $X$~binom. 
Let the $p=0.20$, $q=0.80$, and $n=20$. The cdf is 
$$P(X\geq x)<0.50 = 1 - P(X<x) = 1-P(X\leq x-1) < 0.5$$
Because $X$~binom, we represent $P(X\leq x-1)$ as the finite binomial series
$$1-\sum_{k=0}^{x-1}\left[\binom{20}{k} (0.20)^k (0.80)^{20-k}\right] \tag{1}$$
If we set $x=4$, then we have 
$$1-\sum_{k=0}^{3}\left[\binom{20}{k} (0.20)^k (0.80)^{20-k}\right] \approx 0.588551,$$
which is obviously not smaller than $0.5$.
Setting $x=5$, then we have
$$1-\sum_{k=0}^{4}\left[\binom{20}{k} (0.20)^k (0.80)^{20-k}\right] \approx 0.370352,$$
which satisfies the desired inequality. 
Since this is the first value of $x$ which satisfies our inequality, we take $x=5$ as the number of toasters whose probability of having to be repaired after a year is $50$%.
The second part is similar to the first in its construction. Let $Y$ represent the number of toasters not requiring repairs after a year. Then $p=0.8$ and $q=0.2$. We extend the sample size $n=20$ to our cdf. 
Now,
$$P(Y \geq y) = 1-P(Y < y) = 1-P(Y\leq y-1) > 0.8$$
Our cdf is similar to $(1)$, except the $x's$ are replaced with $y's$, and the $0.20$ and $0.80$ are interchanged.
We are looking for the first value for $y$ such that makes the inequality true where all the previous values for $y$ demonstrated the inequality to be false. 
For instance, when $y=16$, then we have
$$1-\sum_{k=0}^{15} \left[\binom{20}{k} (0.80)^k (0.20)^{20-k} \right] \approx 0.629650,$$
which is not what we require.
Incrementing our value of $y$ to $15$, then we have
$$1-\sum_{k=0}^{14} \left[\binom{20}{k} (0.80)^k (0.20)^{20-k} \right] \approx 0.80421,$$
which is what require.
Notice that the values of $y=1,2,...,14$ all satisfy the inequality, but we're looking for the minimal number of $y$ that satisfies the inequality, which would be the first instance for which the function has a value greater than $0.8.$
Therefore, the number of toasters that do not require repairs after a year has expired is $y=15$.
