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Sav [38]
3 years ago
7

.092 -.15-.19= how do I solve

Mathematics
1 answer:
photoshop1234 [79]3 years ago
5 0
Line up the decimal point to each number in order from left to right and subtract to get the difference which should be -0.248
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f) The life of a power transmission tower is exponentially distributed, with mean life 25 years. If three towers, operated indep
Step2247 [10]

Answer:

15.24% probability that at least 2 will still stand after 35 years

Step-by-step explanation:

To solve this question, we need to understand the binomial distribution and the exponential distribution.

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.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

Probability of a single tower being standing after 35 years:

Single tower, so exponential.

Mean of 25 years, so m = 25, \mu = \frac{1}{25} = 0.04

We have to find P(X > 35)

P(X > 35) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-0.04*35} = 0.2466

What is the probability that at least 2 will still stand after 35 years?

Now binomial.

Each tower has a 0.2466 probability of being standing after 35 years, so p = 0.2466

3 towers, so n = 3

We have to find:

P(X \geq 2) = P(X = 2) + P(X = 3)

In which

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

P(X = 2) = C_{3,2}.(0.2466)^{2}.(0.7534)^{1} = 0.1374

P(X = 3) = C_{3,3}.(0.2466)^{3}.(0.7534)^{0} = 0.0150

P(X \geq 2) = P(X = 2) + P(X = 3) = 0.1374 + 0.0150 = 0.1524

15.24% probability that at least 2 will still stand after 35 years

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