Question

The time , in hours , it takes to repair an electrical breakdown in a certain factory is represent by a random variable x where repair time follow a normal distribution with mean 5 hour and standard deviation 1 hour , if 5 electrical breakdown ore randomly chosen , Find the probability that ( a all repaired time below 6 hours ( b ) exactly 3 of them repaired below 6 hours

Answer 1

Answer:

a) 0.4215 = 42.15% probability that all are repaired in less than 6 hours.

b) 0.15 = 15% probability that exactly 3 of them repaired in a time below 6 hours

Step-by-step explanation:

To solve this question, we need to understand the normal and the binomial probability distributions.

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.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Proportion repaired below 6 hours:

Mean 5 hours and standard deviation 1 hour, which means that $\mu = 5, \sigma = 1$

This proportion is the p-value of Z when X = 6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6 - 5}{1}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

0.8413 repaired below 6 hours.

For the binomial distribution, this means that $p = 0.8413$

5 are chosen:

This means that $n = 5$

a) Probability that all are repaired in less than 6 hours

This is P(X = 5). So

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

$P(X = 5) = C_{5,5}.(0.8413)^{5}.(0.1587)^{0} = 0.4215$

0.4215 = 42.15% probability that all are repaired in less than 6 hours.

b) Probability that exactly 3 of them repaired below 6 hours

This is P(X = 3). So

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

$P(X = 3) = C_{5,3}.(0.8413)^{3}.(0.1587)^{2} = 0.15$

0.15 = 15% probability that exactly 3 of them repaired in a time below 6 hours