Based on a smartphone​ survey, assume that 

4949​% 

of adults with smartphones use them in theaters. In a separate survey of 

226226 

adults with​ smartphones, it is found that 

108108 

use them in theaters.

**a.** If the 

4949​% 

rate is​ correct, find the probability of getting 

108108 

or fewer smartphone owners who use them in theaters.

**b.** Is the result of 

108108 

significantly​ low?

To answer the questions, we will use the normal approximation to the binomial distribution.

### Part (a): Find the probability of getting 108 or fewer smartphone owners who use them in theaters.

#### Step 1: Define the parameters
We have:
- \( n = 226 \): the number of adults surveyed
- \( p = 0.49 \): the proportion of adults with smartphones who use them in theaters

We want to find the probability of getting 108 or fewer smartphone owners who use them in theaters, i.e., \( P(X \leq 108) \).

#### Step 2: Check the conditions for normal approximation
For the normal approximation to be valid, both \( np \) and \( n(1-p) \) should be greater than 5:
- \( np = 226 \times 0.49 = 110.74 \)
- \( n(1-p) = 226 \times 0.51 = 115.26 \)

Both values are greater than 5, so the normal approximation is appropriate.

#### Step 3: Find the mean and standard deviation of the binomial distribution
- Mean \( \mu = np = 226 \times 0.49 = 110.74 \)
- Standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{226 \times 0.49 \times 0.51} \approx 7.51 \)

#### Step 4: Apply the continuity correction
For \( P(X \leq 108) \), we use \( X = 108.5 \).

#### Step 5: Convert to the standard normal distribution
We convert the binomial problem to a standard normal distribution problem using the \( Z \)-score formula:

\[ Z = \frac{X - \mu}{\sigma} \]

For \( X = 108.5 \):
\[ Z = \frac{108.5 - 110.74}{7.51} \approx -0.30 \]

#### Step 6: Find the probability
Using the standard normal distribution table (or a calculator), we find the probability corresponding to \( Z = -0.30 \):

\[ P(Z \leq -0.30) \approx 0.3821 \]

So, the probability of getting 108 or fewer smartphone owners who use them in theaters is approximately 0.3821, or 38.21%.

### Part (b): Is the result of 108 significantly low?

To determine if 108 is significantly low, we can use the concept of a significance level (commonly \( \alpha = 0.05 \) for a 5% significance level). If the probability found in part (a) is less than \( \alpha \), we would consider the result significantly low.

Here, the probability is 0.3821, which is much greater than 0.05. Therefore, the result of 108 smartphone owners using their phones in theaters is not significantly low.

Would you like more details or have any questions?

### Related Questions:
1. What is the probability of getting exactly 108 smartphone owners who use them in theaters?
2. What is the probability of getting more than 120 smartphone owners who use them in theaters?
3. How would the probability change if the percentage was 55% instead of 49%?
4. What is the expected number of smartphone owners who use them in theaters?
5. How does the standard deviation affect the shape of the normal curve in this context?
6. What is the Z-score for 108 smartphone owners?
7. How would the probability change if 300 adults were surveyed instead of 226?
8. How can the normal approximation be validated against the exact binomial probability?

### Tip:
When assessing the significance of a result, compare the calculated probability to the chosen significance level to determine if it is statistically significant.

Explore the probability calculation for smartphone usage in theaters based on survey data. Understand how to use binomial distribution and normal approximation to analyze the findings. Suitable for advanced high school students and beyond.

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