Given a binomial random variable [tex]\(X\)[/tex] with parameters [tex]\(n = 290\)[/tex] and [tex]\(p = 0.66\)[/tex], we will use the normal approximation to find the required probabilities. Here are the steps to solve each part of the problem:
### Step-by-Step Solution:
1. Calculate the mean and standard deviation of the binomial distribution:
The mean ([tex]\(\mu\)[/tex]) of a binomial distribution is given by:
[tex]\[\mu = n \cdot p\][/tex]
The standard deviation ([tex]\(\sigma\)[/tex]) is given by:
[tex]\[\sigma = \sqrt{n \cdot p \cdot (1 - p)}\][/tex]
Plugging in the values:
[tex]\[\mu = 290 \cdot 0.66 = 191.4\][/tex]
[tex]\[\sigma = \sqrt{290 \cdot 0.66 \cdot (1 - 0.66)} \approx 8.067\][/tex]
2. For part (a) [tex]\(P(X \geq 182)\)[/tex]:
We need to use the continuity correction factor because we are approximating a discrete distribution (binomial) with a continuous distribution (normal). For [tex]\(P(X \geq 182)\)[/tex], we use [tex]\(P(X \geq 181.5)\)[/tex].
To find this probability, we convert the binomial variable into a standard normal variable [tex]\(Z\)[/tex] using the z-score formula:
[tex]\[Z = \frac{X - \mu}{\sigma}\][/tex]
For [tex]\(X = 181.5\)[/tex]:
[tex]\[Z = \frac{181.5 - 191.4}{8.067} \approx -1.227\][/tex]
Now, we find the probability:
[tex]\[P(X \geq 182) = P(Z \geq -1.227) = 1 - P(Z \leq -1.227)\][/tex]
Using the standard normal distribution table, [tex]\(P(Z \leq -1.227) \approx 0.1099\)[/tex]:
[tex]\[P(Z \geq -1.227) \approx 1 - 0.1099 \approx 0.8901\][/tex]
Therefore:
[tex]\[P(X \geq 182) \approx 0.8901\][/tex]
3. For part (b) [tex]\(P(X \leq 193)\)[/tex]:
Again, using the continuity correction factor, for [tex]\(P(X \leq 193)\)[/tex], we use [tex]\(P(X \leq 193.5)\)[/tex].
For [tex]\(X = 193.5\)[/tex]:
[tex]\[Z = \frac{193.5 - 191.4}{8.067} \approx 0.260\][/tex]
Now, we find the probability:
[tex]\[P(X \leq 193) = P(Z \leq 0.260)\][/tex]
Using the standard normal distribution table, [tex]\(P(Z \leq 0.260) \approx 0.6027\)[/tex]:
[tex]\[P(X \leq 193) \approx 0.6027\][/tex]
### Summary:
a) [tex]\(P(X \geq 182) \approx 0.8901\)[/tex]
b) [tex]\(P(X \leq 193) \approx 0.6027\)[/tex]
These results give the approximate probabilities for the binomial random variable [tex]\(X\)[/tex] using the normal approximation method.
