One last one.
A poll at Steve's high school was taken to see if students are in favor of spending class money to expand the junior-senior parking lot. Steve surveyed 6 random students from the population. 85% favored the expansion, while 15% opposed it.
a) Determine the probabilites associated with the number of students that Steve asked who are in favor of expanding the parking lot by calculating the probability distribution.
b) What is the probability that no more than 2 people are in favor of expanding the parking lot?
c) How many students should Steve expect to find who are in favor of expanding the parking lot?
To solve this problem, we can use the binomial probability formula. The binomial distribution is used when there are two possible outcomes (in this case, in favor or opposed), and each outcome has a fixed probability of occurring (85% in favor and 15% opposed).
a) To calculate the probability distribution, we need to determine the probability of each possible outcome (0, 1, 2, 3, 4, 5, or 6 students) being in favor of expanding the parking lot. We can use the binomial probability formula:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
Where:
- n is the total number of students surveyed (6 in this case)
- k is the number of students in favor of expanding the parking lot
- p is the probability of a single student being in favor (0.85 in this case)
- (n C k) is the binomial coefficient, calculated as n!/((n-k)! * k!)
Let's calculate the probability distribution:
P(X=0) = (6 C 0) * (0.85^0) * (0.15^6) = 1 * 1 * 0.00000759375 ≈ 0.000007594
P(X=1) = (6 C 1) * (0.85^1) * (0.15^5) = 6 * 0.85 * 0.000759375 ≈ 0.0037396875
P(X=2) = (6 C 2) * (0.85^2) * (0.15^4) = 15 * 0.7225 * 0.0050625 ≈ 0.054140625
P(X=3) = (6 C 3) * (0.85^3) * (0.15^3) = 20 * 0.614125 * 0.003375 ≈ 0.041890625
P(X=4) = (6 C 4) * (0.85^4) * (0.15^2) = 15 * 0.52200625 * 0.00225 ≈ 0.01724140625
P(X=5) = (6 C 5) * (0.85^5) * (0.15^1) = 6 * 0.44320415625 * 0.15 ≈ 0.03978094765
P(X=6) = (6 C 6) * (0.85^6) * (0.15^0) = 1 * 0.376390625 * 1 ≈ 0.376390625
Therefore, the probability distribution of the number of students in favor is approximately:
P(X=0) ≈ 0.000007594
P(X=1) ≈ 0.0037396875
P(X=2) ≈ 0.054140625
P(X=3) ≈ 0.041890625
P(X=4) ≈ 0.01724140625
P(X=5) ≈ 0.03978094765
P(X=6) ≈ 0.376390625
b) To find the probability that no more than 2 people are in favor of expanding the parking lot, we need to sum the probabilities of having 0, 1, or 2 students in favor:
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) ≈ 0.000007594 + 0.0037396875 + 0.054140625 ≈ 0.0588889065
Therefore, the probability that no more than 2 people are in favor of expanding the parking lot is approximately 0.0588889065 or 5.89%.
c) To find the expected number of students in favor, we multiply each possible outcome by its probability and sum them up:
Expected value (E) = (X=0) * P(X=0) + (X=1) * P(X=1) + (X=2) * P(X=2) + ... + (X=6) * P(X=6)
E ≈ (0 * 0.000007594) + (1 * 0.0037396875) + (2 * 0.054140625) + (3 * 0.041890625) + (4 * 0.01724140625) + (5 * 0.03978094765) + (6 * 0.376390625)
E ≈ 0 + 0.0037396875 + 0.10828125 + 0.125671875 + 0.068965625 + 0.19890473825 + 2.25834375
E ≈ 2.764905875
Therefore, Steve should expect to find approximately 2.76 students who are in favor of expanding the parking lot.