2/3 and 5/6 common denominator
1 answer:
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Answer:
a) Figure and code attached
b)
c)
And we can calculate this with the complement rule.
So then we have:
And we are interested on the valueof n who satisfy this expression.
And for this we can verify this with the following code:
"=BINOM.DIST(6,E54,0.2,TRUE)"
And as we can see on the second figure attached the value who satisfy the condition would be n = 60.
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:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Part a
We can use the following R code to generate the histogram for this case:
> x <- seq(0,8,by = 1)
> y <- dbinom(x,8,0.2)
> plot(x,y,type = "h",main="Histogram")
And as we can see we got the result on the figure attached. And the distribution seems to be skewed to the right.
Part b
For this case the expected value is given by:
The variance is given by:
And the standard deviation would be:
Part c
For this case we have the following inequality:
And we can calculate this with the complement rule.
So then we have:
And we are interested on the valueof n who satisfy this expression.
And for this we can verify this with the following code:
"=BINOM.DIST(6,E54,0.2,TRUE)"
And as we can see on the second figure attached the value who satisfy the condition would be n = 60.
