To find the probability that less than 55% of the sampled teenagers own smartphones, we can use the normal approximation to the binomial distribution since the sample size $n = 150$ is large. Here’s how we proceed:

1. Given Values and Setup:

   • Population proportion $p = 0.48$
   • Sample size $n = 150$
   • We are looking for the probability that the sample proportion $\hat{p}$ is less than 0.55.

2. Calculate the Mean and Standard Deviation of the Sampling Distribution:

   • The mean of the sampling distribution for $\hat{p}$ is $\mu_{\hat{p}} = p = 0.48$.
   • The standard deviation of the sampling distribution for $\hat{p}$ is: $\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.48 \times (1 - 0.48)}{150}}.$

   Calculating $\sigma_{\hat{p}}$: $\sigma_{\hat{p}} = \sqrt{\frac{0.48 \times 0.52}{150}} = \sqrt{\frac{0.2496}{150}} \approx 0.0408.$

3. Convert to a Z-Score:

   • We want $P(\hat{p} < 0.55)$, which translates to finding the probability that the sample proportion is less than 0.55.
   • Convert $0.55$ to a z-score using the formula: $z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.55 - 0.48}{0.0408}.$

   Calculating the z-score: $z = \frac{0.55 - 0.48}{0.0408} \approx 1.716.$

4. Find the Probability:

   • Now, we look up the z-score of $1.716$ in the standard normal distribution table or use a calculator. The cumulative probability for $z = 1.716$ is approximately $0.9564$.

Therefore, the probability that less than 55% of the sampled teenagers own smartphones is approximately:

$P(\hat{p} < 0.55) \approx 0.9564.$

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Related Questions

1. How would the probability change if the sample size were increased to 200?
2. What is the probability that more than 60% of sampled teenagers own smartphones?
3. How does the normal approximation apply to smaller sample sizes?
4. If the population proportion were 50%, how would that affect the result?
5. How do we calculate the probability for exactly 55% of teenagers owning smartphones?

Tip:

For problems involving sample proportions, use the normal approximation to the binomial distribution when $np$ and $n(1 - p)$ are both greater than 5.