Question

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 x and y such that (a) the probability that at least x of them will require repairs is less than 0.5; (b) the probability that at least y of them will not require repairs is greater than 0.8

Answer 1

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:

$P(X=x)={20\choose x}(0.20)^{x}(1-0.20)^{20-x}$

$P(Y=y)={20\choose y}(0.20)^{20-y}(1-0.20)^{y}$

(a)

Compute the value of x such that P (X ≥ x) < 0.50:

$P (X \geq x) < 0.50\\\\1-P(X\leq x-1)$

Use the Binomial table for n = 20 and p = 0.20.

$0.411=\sum\limits^{3}_{x=0}[b(x,20,0.20)]$

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:

$P (Y \geq y) >0.80\\\\P(Y\leq 20-y)>0.80\\\\P(Y\leq 20-y)>0.80\\\\\sum\limits^{20-y}_{y=0}[{20\choose y}(0.20)^{20-y}(1-0.20)^{y}]>0.80$

Use the Binomial table for n = 20 and p = 0.20.

$0.630=\sum\limits^{4}_{y=0}[b(y,20,0.20)]$

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.