In a study, 35% of adults questioned reported that their health was excellent. A researcher wishes to study health of people living close to a nuclear power plant. Among 14 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 14 adults are randomly selected, 3 or fewer are in excellent health. SHOW WORK/FORMULA USED OR STATE CALCULATOR STEP USED
To find the probability that 3 or fewer adults out of 14 are in excellent health, we can use the binomial probability formula.
The binomial probability formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
P(x) is the probability of getting exactly x successes
n is the total number of trials
x is the number of successful outcomes
p is the probability of success on a single trial
q is the probability of failure on a single trial, which is equal to 1-p
In this case, we have:
n = 14 (total number of selected adults)
x = 0, 1, 2, 3 (number of adults in excellent health)
p = 0.35 (probability of an adult having excellent health)
q = 1 - p = 1 - 0.35 = 0.65 (probability of an adult not having excellent health)
We need to calculate P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3).
P(x = 0) = (14C0) * 0.35^0 * 0.65^(14-0)
P(x = 0) = 1 * 1 * 0.65^14
P(x = 0) ≈ 0.2868
P(x = 1) = (14C1) * 0.35^1 * 0.65^(14-1)
P(x = 1) = 14 * 0.35 * 0.65^13
P(x = 1) ≈ 0.3952
P(x = 2) = (14C2) * 0.35^2 * 0.65^(14-2)
P(x = 2) = (14 * 13) / (2 * 1) * 0.35^2 * 0.65^12
P(x = 2) ≈ 0.2508
P(x = 3) = (14C3) * 0.35^3 * 0.65^(14-3)
P(x = 3) = (14 * 13 * 12) / (3 * 2 * 1) * 0.35^3 * 0.65^11
P(x = 3) ≈ 0.0926
Now, we calculate the total probability:
P(3 or fewer adults in excellent health) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)
P(3 or fewer adults in excellent health) ≈ 0.2868 + 0.3952 + 0.2508 + 0.0926
P(3 or fewer adults in excellent health) ≈ 0.6254
Therefore, the probability that when 14 adults are randomly selected, 3 or fewer are in excellent health is approximately 0.6254.
To find the probability that 3 or fewer out of 14 adults randomly selected from the area near a nuclear power plant report excellent health, we can use the binomial probability formula:
P(X ≤ 3) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(X ≤ 3) is the probability that X (the number of adults reporting excellent health) is less than or equal to 3,
C(n, x) is the binomial coefficient or the number of ways to choose x successes from n trials,
p is the probability of success in a single trial (the proportion of adults reporting excellent health),
n is the number of trials (the number of adults randomly selected).
In this case, n = 14 (the number of adults randomly selected) and p = 0.35 (the proportion of adults reporting excellent health).
Now, let's calculate the probability:
P(X ≤ 3) = C(14, 0) * 0.35^0 * (1-0.35)^(14-0) +
C(14, 1) * 0.35^1 * (1-0.35)^(14-1) +
C(14, 2) * 0.35^2 * (1-0.35)^(14-2) +
C(14, 3) * 0.35^3 * (1-0.35)^(14-3)
C(14, 0) = 1 (since there is only one way to choose 0 successes from 14 trials)
C(14, 1) = 14 (since there are 14 ways to choose 1 success from 14 trials)
C(14, 2) = 91 (since there are 91 ways to choose 2 successes from 14 trials)
C(14, 3) = 364 (since there are 364 ways to choose 3 successes from 14 trials)
P(X ≤ 3) = 1 * 0.35^0 * 0.65^14 +
14 * 0.35^1 * 0.65^13 +
91 * 0.35^2 * 0.65^12 +
364 * 0.35^3 * 0.65^11
Now, evaluate this expression using a calculator or software to get the final answer.
this is binomial ... excellent health or not
P(e) = .35 ... P(n) = .65
(n + e)^14 ... sum the 1st 4 terms
n^14 + 14 n^13 e + 91 n^12 e^2 + 364 n^11 e^3
.65^14 + (14 * .65^13 * .35) + ...