To solve the given problem, we need to approach it using the binomial distribution and the normal approximation to the binomial distribution. Here are the detailed steps for each part:

Part (a): Probability of Getting 108 or Fewer Smartphone Owners

1. Define the Parameters:

   • $n = 226$ (total number of adults surveyed)
   • $p = 0.49$ (probability of using a smartphone in theaters)
   • $x = 108$ (number of adults who use smartphones in theaters)

2. Calculate the Mean and Standard Deviation:

   • Mean ($\mu$) of the binomial distribution: $\mu = n \cdot p = 226 \cdot 0.49 = 110.74$
   • Standard deviation ($\sigma$) of the binomial distribution: $\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{226 \cdot 0.49 \cdot 0.51} \approx 7.516$

3. Use the Normal Approximation: Since $n$ is large and $p$ is not too close to 0 or 1, we can use the normal approximation to the binomial distribution.

4. Convert $x = 108$ to a Z-Score: $Z = \frac{x - \mu}{\sigma} = \frac{108 - 110.74}{7.516} \approx -0.365$

5. Find the Probability Using the Z-Score: We need to find the cumulative probability up to $Z = -0.365$.

   Using the standard normal distribution table or a calculator, we find: $P(Z \leq -0.365) \approx 0.3577$

   So, the probability of getting 108 or fewer smartphone owners who use them in theaters is approximately 0.3577.

Part (b): Significance of the Result

To determine if 108 is significantly low, we can use the concept of significance levels and critical values.

1. Common Significance Level: A common significance level is 0.05. We compare the probability we found in part (a) to this significance level.

2. Interpret the Probability: The probability of getting 108 or fewer smartphone users in theaters, assuming the true rate is 49%, is approximately 0.3577. Since 0.3577 is much larger than 0.05, the result of 108 is not considered significantly low.

Summary

Part (a): The probability of getting 108 or fewer smartphone owners who use them in theaters is approximately 0.3577.

Part (b): The result of 108 is not significantly low because the probability (0.3577) is much larger than the common significance level of 0.05.

Would you like more details or have any questions?

Here are 8 related questions to explore further:

1. How do you calculate the exact binomial probability for $P(X \leq 108)$ without using normal approximation?
2. What is the critical value for a significance level of 0.05 in a normal distribution?
3. How would the probability change if the sample size were increased to 500?
4. What is the concept of a p-value in hypothesis testing?
5. How does the shape of the binomial distribution change as $n$ increases?
6. Can you explain the central limit theorem and its relevance to normal approximation?
7. How would you determine if a different sample proportion (e.g., 55%) is significantly different from 49%?
8. What are Type I and Type II errors in the context of hypothesis testing?

Tip: Always check if the conditions for normal approximation to the binomial distribution are met before applying it: $np$ and $n(1-p)$ should both be greater than 5.