Answer:
36.58% probability that one of the devices fail
Step-by-step explanation:
For each device, there are only two possible outcomes. Either it fails, or it does not fail. The probability of a device failling is independent of other devices. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

And p is the probability of X happening.
A total of 15 devices will be used.
This means that 
Assume that each device has a probability of 0.05 of failure during the course of the monitoring period.
This means that 
What is the probability that one of the devices fail?
This is 


36.58% probability that one of the devices fail
Ok name the expressions we both know 3z is one 5x is one 6xy is one do you think 3x^2y^4z is an expression i would give you the answer but i also want you to understand the problem and what your looking for to in the equation
Answer:
8c proved by me
Step-by-step explanation:
answer is 8c proved
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Answer:
(a) 0.40
(b) 0.049
(c) 
(d) Explained below
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:

The standard deviation of this sampling distribution of sample proportion is:

Given:
n = 100
p = 0.40
As <em>n</em> = 100 > 30 the Central limit theorem is applicable.
(a)
Compute the expected value of
as follows:

The expected value of
is 0.40.
(b)
Compute the standard error of
as follows:

The standard error of
is 0.049.
(c)
The sampling distribution of
is:

(d)
The sampling distribution of p show that as the sample size is increasing the distribution is approximated by the normal distribution.