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Sure, I can guide you through this step-by-step process to find the probability that more than 82% of the sampled teenagers own a smartphone.
We have the following information:
- The proportion of teenagers owning smartphones, [tex]\( p = 0.80 \)[/tex]
- The sample size, [tex]\( n = 250 \)[/tex]
### Part 1: Find the mean [tex]\(\mu_p\)[/tex].
The mean [tex]\(\mu_p\)[/tex] of the sampling distribution is given by the population proportion:
[tex]\[ \mu_p = p = 0.8000 \][/tex]
### Part 2: Find the standard deviation [tex]\(\sigma_{\hat{p}}\)[/tex].
The standard deviation [tex]\(\sigma_{\hat{p}}\)[/tex] (also known as the standard error) of the sample proportion is calculated using the formula:
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \][/tex]
Substituting the given values,
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{0.80 \times (1 - 0.80)}{250}} = \sqrt{\frac{0.80 \times 0.20}{250}} = \sqrt{\frac{0.16}{250}} = \sqrt{0.00064} = 0.0253 \][/tex]
### Part 3: Find the probability that more than 82% of the sampled teenagers own a smartphone.
We need to find the probability that the sample proportion ([tex]\(\hat{p}\)[/tex]) is greater than 0.82.
To find this probability, we first convert the sample proportion to a z-score using the z-formula:
[tex]\[ z = \frac{\hat{p} - \mu_p}{\sigma_{\hat{p}}} \][/tex]
For [tex]\(\hat{p} = 0.82\)[/tex],
[tex]\[ z = \frac{0.82 - 0.80}{0.0253} = \frac{0.02}{0.0253} \approx 0.7906 \][/tex]
Next, we use the standard normal distribution to find the probability corresponding to this z-score. The cumulative distribution function (CDF) of the normal distribution gives us the probability that a value is less than a given z-score.
Using this probability (from standard normal distribution tables or computational tools), we find:
[tex]\[ P(Z \leq 0.7906) \approx 0.7854 \][/tex]
Since we want the probability that the proportion is more than 0.82, we subtract this value from 1:
[tex]\[ P(\hat{p} > 0.82) = 1 - P(Z \leq 0.7906) = 1 - 0.7854 \approx 0.2146 \][/tex]
Thus, the probability that more than 82% of the sampled teenagers own a smartphone is approximately:
[tex]\[ \boxed{0.2146} \][/tex]
We have the following information:
- The proportion of teenagers owning smartphones, [tex]\( p = 0.80 \)[/tex]
- The sample size, [tex]\( n = 250 \)[/tex]
### Part 1: Find the mean [tex]\(\mu_p\)[/tex].
The mean [tex]\(\mu_p\)[/tex] of the sampling distribution is given by the population proportion:
[tex]\[ \mu_p = p = 0.8000 \][/tex]
### Part 2: Find the standard deviation [tex]\(\sigma_{\hat{p}}\)[/tex].
The standard deviation [tex]\(\sigma_{\hat{p}}\)[/tex] (also known as the standard error) of the sample proportion is calculated using the formula:
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \][/tex]
Substituting the given values,
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{0.80 \times (1 - 0.80)}{250}} = \sqrt{\frac{0.80 \times 0.20}{250}} = \sqrt{\frac{0.16}{250}} = \sqrt{0.00064} = 0.0253 \][/tex]
### Part 3: Find the probability that more than 82% of the sampled teenagers own a smartphone.
We need to find the probability that the sample proportion ([tex]\(\hat{p}\)[/tex]) is greater than 0.82.
To find this probability, we first convert the sample proportion to a z-score using the z-formula:
[tex]\[ z = \frac{\hat{p} - \mu_p}{\sigma_{\hat{p}}} \][/tex]
For [tex]\(\hat{p} = 0.82\)[/tex],
[tex]\[ z = \frac{0.82 - 0.80}{0.0253} = \frac{0.02}{0.0253} \approx 0.7906 \][/tex]
Next, we use the standard normal distribution to find the probability corresponding to this z-score. The cumulative distribution function (CDF) of the normal distribution gives us the probability that a value is less than a given z-score.
Using this probability (from standard normal distribution tables or computational tools), we find:
[tex]\[ P(Z \leq 0.7906) \approx 0.7854 \][/tex]
Since we want the probability that the proportion is more than 0.82, we subtract this value from 1:
[tex]\[ P(\hat{p} > 0.82) = 1 - P(Z \leq 0.7906) = 1 - 0.7854 \approx 0.2146 \][/tex]
Thus, the probability that more than 82% of the sampled teenagers own a smartphone is approximately:
[tex]\[ \boxed{0.2146} \][/tex]
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