Answer:
(a) The value of x is 5.
(b) The value of y is 15.
Step-by-step explanation:
Let the random variable X represent the number of electric toasters produced that require repairs within 1 year.
And the let the random variable Y 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 n = 20 toasters are selected.
The random variable X and Y thus, follows binomial distribution.
The probability mass function of X and Y are:
[tex]P(X=x)={20\choose x}(0.20)^{x}(1-0.20)^{20-x}[/tex]
[tex]P(Y=y)={20\choose y}(0.20)^{20-y}(1-0.20)^{y}[/tex]
(a)
Compute the value of x such that P (X ≥ x) < 0.50:
[tex]P (X \geq x) < 0.50\\\\1-P(X\leq x-1)<0.50\\\\0.50<P(X\leq x-1)\\\\0.50<\sum\limits^{x-1}_{0}[{20\choose x}(0.20)^{x}(1-0.20)^{20-x}][/tex]
Use the Binomial table for n = 20 and p = 0.20.
[tex]0.411=\sum\limits^{3}_{x=0}[b(x,20,0.20)]<0.50<\sum\limits^{4}_{x=0}[b(x,20,0.20)]=0.630[/tex]
The least value of x that satisfies the inequality P (X ≥ x) < 0.50 is:
x - 1 = 4
x = 5
Thus, the value of x is 5.
(b)
Compute the value of y such that P (Y ≥ y) > 0.80:
[tex]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[/tex]
Use the Binomial table for n = 20 and p = 0.20.
[tex]0.630=\sum\limits^{4}_{y=0}[b(y,20,0.20)]<0.50<\sum\limits^{5}_{y=0}[b(y,20,0.20)]=0.804[/tex]
The least value of y that satisfies the inequality P (Y ≥ y) > 0.80 is:
20 - y = 5
y = 15
Thus, the value of y is 15.