Answer :
Answer:
Let the digits from 0 to 3 represent birds with a wingspan greater than 10 inches and the remaining digits represent birds with a wingspan less than or equal to 10 inches.
Step-by-step explanation:
The choices of the question is missing. Choices are:
a.Let the even digits represent birds with a wingspan greater than 10 inches and the odd digits represent birds with a wingspan less than or equal to 10 inches.
b.Let the digits 0 and 1 represent birds with a wingspan greater than 10 inches and the remaining digits represent birds with a wingspan less than or equal to 10 inches.
c.Let the digits from 0 to 2 represent birds with a wingspan greater than 10 inches and the remaining digits represent birds with a wingspan less than or equal to 10 inches.
d.Let the digits from 0 to 3 represent birds with a wingspan greater than 10 inches and the remaining digits represent birds with a wingspan less than or equal to 10 inches.
e.Let the digits from 0 to 4 represent birds with a wingspan greater than 10 inches and the remaining digits represent birds with a wingspan less than or equal to 10 inches.
The population proportion of birds which have a wingspan greater than 10 inches is 40% (0.40)
This proportion need to be reflected in the assignments of the digits 0 to 9 in order to model the population.
If the digits from 0 to 3 represent birds with a wingspan greater than 10 inches and the remaining digits represent birds with a wingspan less than or equal to 10 inches, this proporion is satisfied because 4 digits are assigned to birds with a wingspan greater than 10 inches and 6 digits are assigned to birds with a wingspan less than or equal to 10 inches.
Using the binomial distribution, it is found that there is a 0.1866 = 18.66% probability of finding 1 bird in a sample of 6 birds with a wingspan greater than 10 inches.
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For each bird, there are only two possible outcomes. Either they have a wingspan greater than 10 inches, or they do not. The probability of a bird having a wingspan greater than 10 inches is independent of any other bird, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
p is the probability of a success on a single trial.
In this problem:
- 40% have a wingspan greater than 10 inches, thus [tex]p = 0.4[/tex]
- Sample of 6, thus [tex]n = 6[/tex]
The probability of finding one with a wingspan greater than 10 inches is P(X = 1), thus:
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{6,1}.(0.4)^{1}.(0.6)^{5} = 0.1866[/tex]
0.1866 = 18.66% probability of finding 1 bird in a sample of 6 birds with a wingspan greater than 10 inches.
A similar problem is given at https://brainly.com/question/24863377