Answer:
(a) The value of <em>x</em> is 5.
(b) The value of <em>y</em> is 15.
Step-by-step explanation:
Let the random variable <em>X</em> represent the number of electric toasters produced that require repairs within 1 year.
And the let the random variable <em>Y</em> represent the number of electric toasters produced that does not require repairs within 1 year.
The probability of the random variables are:
P (X) = 0.20
P (Y) = 1 - P (X) = 1 - 0.20 = 0.80
The event that a randomly selected electric toaster requires repair is independent of the other electric toasters.
A random sample of <em>n</em> = 20 toasters are selected.
The random variable <em>X</em> and <em>Y</em> thus, follows binomial distribution.
The probability mass function of <em>X</em> and <em>Y</em> are:


(a)
Compute the value of <em>x</em> such that P (X ≥ x) < 0.50:

Use the Binomial table for <em>n</em> = 20 and <em>p</em> = 0.20.
![0.411=\sum\limits^{3}_{x=0}[b(x,20,0.20)]](https://tex.z-dn.net/?f=0.411%3D%5Csum%5Climits%5E%7B3%7D_%7Bx%3D0%7D%5Bb%28x%2C20%2C0.20%29%5D%3C0.50%3C%5Csum%5Climits%5E%7B4%7D_%7Bx%3D0%7D%5Bb%28x%2C20%2C0.20%29%5D%3D0.630)
The least value of <em>x</em> that satisfies the inequality P (X ≥ x) < 0.50 is:
<em>x</em> - 1 = 4
<em>x</em> = 5
Thus, the value of <em>x</em> is 5.
(b)
Compute the value of <em>y</em> such that P (Y ≥ y) > 0.80:
![P (Y \geq y) >0.80\\\\P(Y\leq 20-y)>0.80\\\\P(Y\leq 20-y)>0.80\\\\\sum\limits^{20-y}_{y=0}[{20\choose y}(0.20)^{20-y}(1-0.20)^{y}]>0.80](https://tex.z-dn.net/?f=P%20%28Y%20%5Cgeq%20y%29%20%3E0.80%5C%5C%5C%5CP%28Y%5Cleq%2020-y%29%3E0.80%5C%5C%5C%5CP%28Y%5Cleq%2020-y%29%3E0.80%5C%5C%5C%5C%5Csum%5Climits%5E%7B20-y%7D_%7By%3D0%7D%5B%7B20%5Cchoose%20y%7D%280.20%29%5E%7B20-y%7D%281-0.20%29%5E%7By%7D%5D%3E0.80)
Use the Binomial table for <em>n</em> = 20 and <em>p</em> = 0.20.
![0.630=\sum\limits^{4}_{y=0}[b(y,20,0.20)]](https://tex.z-dn.net/?f=0.630%3D%5Csum%5Climits%5E%7B4%7D_%7By%3D0%7D%5Bb%28y%2C20%2C0.20%29%5D%3C0.50%3C%5Csum%5Climits%5E%7B5%7D_%7By%3D0%7D%5Bb%28y%2C20%2C0.20%29%5D%3D0.804)
The least value of <em>y</em> that satisfies the inequality P (Y ≥ y) > 0.80 is:
20 <em>- y</em> = 5
<em>y</em> = 15
Thus, the value of <em>y</em> is 15.