Respuesta :
Answer:
a) P=0.1721
b) P=0.3528
c) P=0.3981
Step-by-step explanation:
This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.
a) In this case we have a sample size of n=15.
The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.
The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.
This probability can be calculated as:
[tex]P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721[/tex]
b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.
We apply the same formula but as a sum:
[tex]P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}[/tex]
Then we have:
[tex]P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528[/tex]
c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).
The first (k less or equal to 5) is already calculated.
We have to calculate for k equal to 10 or more.
[tex]P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}[/tex]
Then we have
[tex]P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453[/tex]
The sum of the probabilities is
[tex]P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981[/tex]
