To solve this problem, we can use the normal approximation to the binomial distribution. Since we're dealing with a large sample size, this approximation will help us simplify the calculation.

Here’s the process:

Step 1: Identify the parameters of the binomial distribution

• Number of trials, $n = 156$ (total people).
• Probability of success (i.e., a person not getting the flu), $p = 0.64$.
• The number of successes we want, $k = 90$.

Step 2: Compute the mean and standard deviation of the binomial distribution

The mean $\mu$ and standard deviation $\sigma$ for a binomial distribution are calculated as follows:

• $\mu = n \cdot p = 156 \cdot 0.64 = 99.84$
• $\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{156 \cdot 0.64 \cdot (1 - 0.64)} = \sqrt{156 \cdot 0.64 \cdot 0.36} \approx 6.0561$

Step 3: Apply the continuity correction

Since we are approximating a binomial distribution with a continuous normal distribution, we apply a continuity correction. This involves adjusting the target number $k = 90$ by 0.5 in both directions.

Thus, we want the probability that the number of people not getting the flu is between $89.5$ and $90.5$.

Step 4: Convert to the standard normal distribution (Z-scores)

Now, we convert the values $89.5$ and $90.5$ to Z-scores using the formula: $Z = \frac{X - \mu}{\sigma}$

• For $X = 89.5$: $Z_1 = \frac{89.5 - 99.84}{6.0561} \approx -1.707$
• For $X = 90.5$: $Z_2 = \frac{90.5 - 99.84}{6.0561} \approx -1.543$

Step 5: Find the probabilities from the Z-table

Using the standard normal table or a calculator:

• The probability corresponding to $Z_1 = -1.707$ is approximately $P(Z_1) = 0.0438$.
• The probability corresponding to $Z_2 = -1.543$ is approximately $P(Z_2) = 0.0617$.

Step 6: Compute the probability

The probability that exactly 90 out of 156 people will not get the flu is the difference between these two probabilities: $P(89.5 < X < 90.5) = P(Z_2) - P(Z_1) = 0.0617 - 0.0438 = 0.0179$

Final Answer:

The approximate probability is $0.0179$, or 1.79%.

Would you like more details on any step or have questions?

Here are five related questions to explore further:

1. How would the result change if the probability of not getting the flu was 70%?
2. Can you apply normal approximation when the sample size is very small?
3. What is the significance of using continuity correction in normal approximation?
4. How does the standard deviation affect the normal approximation?
5. What are the limitations of using the normal approximation for binomial probabilities?

Tip: When using the normal approximation to the binomial, always apply continuity correction for better accuracy.