The percentage of adults who have been told at some point that they have hyptertension is 23.53%. If 10 adults are randomly selected, what is the probability that at least 8 have been told that they have hypertension?
gang
Well, it seems we've stumbled into a mathematical territory. Don't worry, I won't calculate this with clown math. Let's break it down logically, shall we?
Out of the 10 randomly selected adults, we want to find the probability that at least 8 have been told they have hypertension.
To calculate this, we'll calculate the probability of exactly 8 adults having hypertension, then the probability of exactly 9 adults having hypertension, and finally the probability of exactly 10 adults having hypertension.
Since the probability of an adult being told they have hypertension is 23.53%, let's call it P_hyp. The probability of an adult not being told they have hypertension, P_no_hyp, would be 100% minus P_hyp.
Using these probabilities, we can calculate the probability of exactly k adults having hypertension as P_hyp^k * P_no_hyp^(10 - k).
Now, let's add up those probabilities for k = 8, 9, and 10 to find the probability that at least 8 adults have been told that they have hypertension.
Probability of exactly 8 adults having hypertension: P_hyp^8 * P_no_hyp^2
Probability of exactly 9 adults having hypertension: P_hyp^9 * P_no_hyp^1
Probability of exactly 10 adults having hypertension: P_hyp^10 * P_no_hyp^0 (which is just P_hyp^10)
Finally, we'll sum up these three probabilities: P(k>=8) = P(8) + P(9) + P(10).
Now, it's time for some serious calculation!
To find the probability that at least 8 adults have been told they have hypertension out of a random selection of 10 adults, we can use the binomial probability formula.
The binomial probability formula is:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
P(X = k) is the probability of getting exactly k successes
C(n, k) is the number of combinations of n objects taken k at a time (n choose k)
p is the probability of success on a single trial
(1-p) is the probability of failure on a single trial
n is the total number of trials
In this case, the probability of an adult being told they have hypertension is 23.53% or 0.2353. Since we want at least 8 adults out of the 10 to be told they have hypertension, the probability can be calculated as follows:
P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10)
P(X = 8) = C(10, 8) * 0.2353^8 * (1 - 0.2353)^(10-8)
P(X = 9) = C(10, 9) * 0.2353^9 * (1 - 0.2353)^(10-9)
P(X = 10) = C(10, 10) * 0.2353^10 * (1 - 0.2353)^(10-10)
Using a combination calculator, we can determine the values of C(10, 8), C(10, 9), and C(10, 10):
C(10, 8) = 45
C(10, 9) = 10
C(10, 10) = 1
Plugging in these values into the respective probability calculations:
P(X = 8) = 45 * 0.2353^8 * (1 - 0.2353)^(10-8)
P(X = 9) = 10 * 0.2353^9 * (1 - 0.2353)^(10-9)
P(X = 10) = 1 * 0.2353^10 * (1 - 0.2353)^(10-10)
Finally, we can calculate the probability that at least 8 adults have been told they have hypertension:
P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10)
Note: P(X = 10) will be very small because it means all 10 adults have hypertension.
P(X >= 8) = (45 * 0.2353^8 * (1 - 0.2353)^(10-8)) + (10 * 0.2353^9 * (1 - 0.2353)^(10-9)) + (1 * 0.2353^10 * (1 - 0.2353)^(10-10))
You can use a calculator to evaluate this expression and get the final probability value.
To find the probability that at least 8 out of 10 randomly selected adults have been told they have hypertension, we need to calculate the probability of having 8, 9, or 10 adults with hypertension, and then sum up these probabilities.
Let's start by calculating the probability of having exactly 8 adults with hypertension:
P(8 adults with hypertension) = C(10, 8) * (0.2353)^8 * (1 - 0.2353)^(10 - 8)
In this equation, C(10, 8) represents the number of ways to choose 8 individuals out of 10, which can be calculated using the combination formula:
C(n, r) = n! / (r! * (n - r)!)
where n is the total number of individuals (10) and r is the number of individuals chosen (8).
Next, (0.2353)^8 represents the probability that a randomly selected adult has hypertension, raised to the power of 8, because we want 8 adults with hypertension.
(1 - 0.2353)^(10 - 8) represents the probability that a randomly selected adult does not have hypertension, raised to the power of 2, because we want 2 adults without hypertension.
Now, calculate the probabilities for having exactly 9 or 10 adults with hypertension in a similar manner:
P(9 adults with hypertension) = C(10, 9) * (0.2353)^9 * (1 - 0.2353)^(10 - 9)
P(10 adults with hypertension) = C(10, 10) * (0.2353)^10 * (1 - 0.2353)^(10 - 10)
Finally, sum up these three probabilities to find the probability of having at least 8 adults with hypertension:
P(at least 8 adults with hypertension) = P(8 adults with hypertension) + P(9 adults with hypertension) + P(10 adults with hypertension)
By calculating these probabilities and adding them together, you can find the overall probability.
To find the probability that 8 have hypertension is .2353 to the 8th power. However, since it is "at least 8", it must also include 9 and 10 (.2353^9 and .2353^10).
To find the probability of either 8, 9 or 10, add the three products.
I hope this helps. Thanks for asking.