Question

One percent of a certain model of television have defective speakers. Suppose 500 televisions of this model are ready to ship. Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers (round off to second decimal place).

Answer 1

Answer:

49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers

Step-by-step explanation:

For each television, there are only two possible outcomes. Either they have defective speakers, or they do not. The probabilities of each television having defective speakers are independent from each other. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.01, n = 500$

Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers

This is

$P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{500,5}.(0.01)^{5}.(0.99)^{495} = 0.1764$

$P(X = 6) = C_{500,6}.(0.01)^{6}.(0.99)^{494} = 0.1470$

$P(X = 7) = C_{500,7}.(0.01)^{7}.(0.99)^{493} = 0.1048$

$P(X = 8) = C_{500,8}.(0.01)^{8}.(0.99)^{492} = 0.0652$

$P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1764 + 0.1470 + 0.1048 + 0.0652 = 0.4934$

49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers