Question

A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates without failure during the useful life of the product is 0.995, and assume that the components fail independently. Approximate the probability that 5 or more of the original 2000 components fail during the useful life of the product.
(apply the ±½ correction factor and round value of standard normal random variable to 2 decimal places.) Round your answer to four decimal places (e.g. 98.7654).

Answer 1

Answer:

$P(\geg 5)= P(X > 4.5)$

We can use the z score formula given by:

$z= \frac{x-\mu}{\sigma}$

And using this formula we have:

$P(X>4.5) = P(Z>\frac{4.5-10}{0.224}) =P(Z>-24.60)$

And using the complement rule we have this:

$P(Z>24.60)= 1-P(Z<-24.60) = 1-0.9999 \approx 0.0001$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=2000, p=1-0.995= 0.005)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

And we want to find this probability:

$P(X\geq 5)$

For this case since we have independence and the following two conditions:

$np = 2000*0.005 = 10 \geq 10[/tx]

[tex] n(1-p) = 2000*(1-0.005)= 1990 \geq 10$

We can use the normal approximation from the binomial distribution, and the mean would be given by:

$\mu = np =2000*0.005= 10$

$\sigma = \sqrt{np(1-p)}= \sqrt{2000*0.005(1-0.005)}= 0.224$

And using the continuity correction factor we have this:

$P(\geg 5)= P(X > 4.5)$

We can use the z score formula given by:

$z= \frac{x-\mu}{\sigma}$

And using this formula we have:

$P(X>4.5) = P(Z>\frac{4.5-10}{0.224}) =P(Z>-24.60)$

And using the complement rule we have this:

$P(Z>24.60)= 1-P(Z<-24.60) = 1-0.9999 \approx 0.0001$