Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.