Questions: According to a recent survey, 74% of teens ages 12-17 in a certain country used social networks in 2009. A random sample of 140 teenagers from this age group was selected. Complete parts a through d below. a. Calculate the standard error of the proportion. σp=0.0371 (Round to four decimal places as needed.) b. What is the probability that less than 76% of the teens from this sample used social networks? P (Less than 76% of the teens from this sample used social networks) = (Round to four decimal places as needed.)

Solution Steps

The standard error of the proportion is calculated using the formula:

\[ \sigma_{p} = \sqrt{\frac{p(1 - p)}{n}} \]

Substituting the values \( p = 0.74 \) and \( n = 140 \):

\[ \sigma_{p} = \sqrt{\frac{0.74(1 - 0.74)}{140}} = \sqrt{\frac{0.74 \times 0.26}{140}} \approx 0.0371 \]

Thus, the standard error is:

\[ \boxed{\sigma_{p} = 0.0371} \]

To find the probability that less than \( 76\% \) of the teens used social networks, we need to calculate:

\[ P(p < 0.76) \]

This can be expressed in terms of the cumulative distribution function \( \Phi \):

\[ P(p < 0.76) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

Where \( Z_{end} \) is the Z-score corresponding to \( p = 0.76 \) and \( Z_{start} \) approaches negative infinity. The Z-score is calculated as:

\[ Z = \frac{p - \mu}{\sigma} = \frac{0.76 - 0.74}{0.0371} \approx 6.3835 \]

Since \( \Phi(Z_{start}) \) approaches \( 0 \) as \( Z_{start} \) approaches negative infinity, we have:

\[ P(p < 0.76) = \Phi(6.3835) - 0 \approx 1.0 \]

Thus, the probability that less than \( 76\% \) of the teens used social networks is:

\[ \boxed{P < 0.76 = 1.0000} \]

Final Answer