Answer:
0.25ab
Step-by-step explanation:
Data provided in the question:
f(x) = xa(1−x)b, 0≤x≤1
or
f(x) = ab(x−x²)
for point of maxima and minima put f'(x) = 0
Thus,
f'(x) = ab(1 - 2x) = 0
or
1 - 2x = 0
or
x = = 0.5
Now,
to check the condition of maxima or minima
f''(x) = ab(0 - 2) = -2ab
since,
f''(x) < 0
therefore,
x = 0.5 is point of maxima
and the maximum value at x = 0.5 of the function is
f(0.5) = ab(0.5 - 0.5²)
= ab(0.25)
= 0.25ab
Using the binomial distribution, it is found that there is a 0.0555 = 5.55% probability that more than 3 of these selected children have three siblings.
<h3>What is the binomial distribution formula?</h3>The formula is:
The parameters are:
In this problem, for the parameters, we consider that:
The probability that more than 3 of these selected children have three siblings is given by:
In which:
Then:
P(X > 3) = 1 - 0.9445 = 0.0555
0.0555 = 5.55% probability that more than 3 of these selected children have three siblings.
More can be learned about the binomial distribution at brainly.com/question/24863377