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

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

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Kim Chi 5 months 2021-09-01T03:17:58+00:00 1 Answers 3 views 0

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    2021-09-01T03:19:41+00:00

    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

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