To solve these problems, we will use the binomial probability formula, since we're dealing with a fixed number of trials (in this case, 23 adults), each having two possible outcomes (having hypertension or not).
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
where:
- \( n \) = number of trials (23 in this case),
- \( k \) = number of successes (the number of adults with hypertension),
- \( p \) = probability of success (0.33 for hypertension),
- \( \binom{n}{k} \) = binomial coefficient.
Let's compute the required probabilities step by step.
(a) Probability that exactly 6 have hypertension:
Here, \( n = 23 \), \( k = 6 \), and \( p = 0.33 \).
\[ P(X = 6) = \binom{23}{6} (0.33)^6 (0.67)^{23 - 6} \]
Calculating the binomial coefficient:
\[ \binom{23}{6} = \frac{23!}{6!(23 - 6)!} = \frac{23!}{6! \cdot 17!} = 12376 \]
Now substituting the values into the formula:
\[ P(X = 6) = 12376 \times (0.33)^6 \times (0.67)^{17} \]
Calculating \( (0.33)^6 \) and \( (0.67)^{17} \):
\[ (0.33)^6 \approx 0.004833 \] \[ (0.67)^{17} \approx 0.104430 \]
Now compute the probability:
\[ P(X = 6) = 12376 \times 0.004833 \times 0.104430 \approx 0.0613 \]
Thus, the probability that exactly 6 of them have hypertension is approximately 0.0613.
(b) Probability that more than 8 have hypertension:
To find this, we calculate:
\[ P(X > 8) = 1 - P(X \leq 8) = 1 - \sum_{k=0}^{8} P(X = k) \]
We can calculate \( P(X = k) \) for \( k = 0, 1, ..., 8 \) using the same binomial formula and then sum the results.
Alternatively, we could use a binomial cumulative distribution function calculator for convenience, but for simplicity, I will compute a few probabilities directly for \( k \) from 0 to 8:
Using a calculator or binomial distribution table for \( n = 23 \) and \( p = 0.33 \) gives us:
After computing or looking up, we find:
\[ P(X \leq 8) \approx 0.8293 \]
Then, we have:
\[ P(X > 8) = 1 - P(X \leq 8) \approx 0.1707 \]
Thus, the probability that more than 8 have hypertension is approximately 0.1707.
(c) Probability that fewer than 3 have hypertension:
This can be found similarly to before:
\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]
Calculating these:
For \( k=0 \):
\[ P(X = 0) = \binom{23}{0} (0.33)^0 (0.67)^{23} = 1 \cdot 1 \cdot (0.67)^{23} \approx 0.0042 \]
For \( k=1 \):
\[ P(X = 1) = \binom{23}{1} (0.33)^1 (0.67)^{22} = 23 \cdot 0.33 \cdot (0.67)^{22} \approx 0.0410 \]
For \( k=2 \):
\[ P(X = 2) = \binom{23}{2} (0.33)^2 (0.67)^{21} = 253 \cdot (0.33)^2 \cdot (0.67)^{21} \approx 0.1312 \]
Now summing these probabilities gives us:
\[ P(X < 3) \approx 0.0042 + 0.0410 + 0.1312 \approx 0.1764 \]
Thus, the probability that fewer than 3 have hypertension is approximately 0.1764.
(d) Would it be unusual if more than 8 of them have hypertension?
Generally, an event is considered unusual if its probability is less than 0.05 (5%).
From part (b), we found that:
\[ P(X > 8) \approx 0.1707 \]
Since \( 0.1707 > 0.05 \), it would not be unusual for more than 8 of them to have hypertension.
Thus, the complete response for part (d):
It would not be unusual if more than 8 of them have hypertension since the probability is 0.1707.
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