This problem involves calculating the probability of observing 134 or fewer adults using smartphones in theaters given that the true population proportion is 52%. We will solve it using the normal approximation to the binomial distribution.

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Part (a): Find the Probability of Getting 134 or Fewer Smartphone Users

1. Define the variables

• Sample size: $n = 264$
• Proportion of smartphone users who use them in theaters: $p = 0.52$
• Complement: $q = 1 - p = 0.48$
• Observed number: $X = 134$

2. Find the mean and standard deviation

Using the formulas for the mean $\mu$ and standard deviation $\sigma$ of a binomial distribution:

$\mu = n \cdot p = 264 \cdot 0.52 = 137.28$

$\sigma = \sqrt{n \cdot p \cdot q} = \sqrt{264 \cdot 0.52 \cdot 0.48} = \sqrt{65.8944} \approx 8.12$

3. Apply the normal approximation to the binomial distribution

We can approximate the binomial distribution using a normal distribution with mean $\mu$ and standard deviation $\sigma$. We also apply the continuity correction by subtracting 0.5 from 134 to account for the discrete nature of the binomial distribution.

Thus, we need to find:

$P(X \leq 134) = P\left(Z \leq \frac{134 - \mu}{\sigma}\right)$

4. Standardize the value using the Z-score formula

$Z = \frac{134 - 137.28}{8.12} = \frac{-3.28}{8.12} \approx -0.40$

5. Find the probability from the Z-table

Using a Z-table, we look up the probability corresponding to $Z = -0.40$.

$P(Z \leq -0.40) \approx 0.3446$

So, the probability of getting 134 or fewer smartphone users who use them in theaters is approximately:

$P(X \leq 134) \approx 0.3446$

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Part (b): Is the result of 134 significantly low?

To determine if 134 is significantly low, we compare its Z-score with typical significance levels. Common thresholds are Z = -1.96 for a 5% significance level (two-tailed) or Z = -1.645 for a 5% significance level (one-tailed).

• Z-score we found: $Z \approx -0.40$

Since $-0.40$ is not below these critical Z-scores, 134 is not significantly low. There is a reasonable chance (34.46%) of getting 134 or fewer smartphone users, assuming the true proportion is 52%.

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Final Answer

(a) The probability of getting 134 or fewer smartphone users who use them in theaters is approximately 0.3446.
(b) No, the result of 134 is not significantly low.

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Would you like more details or have any follow-up questions?

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Here are 5 related questions you might find interesting:

1. How does the continuity correction affect normal approximation in binomial distributions?
2. What is the difference between a one-tailed and two-tailed Z-test?
3. How would the answer change if the proportion was 60% instead of 52%?
4. When should we use the normal approximation to the binomial distribution?
5. How do changes in sample size affect the mean and standard deviation in binomial distributions?

Tip: If the sample size is large and the success probability is close to 0.5, the normal approximation becomes more accurate.