8. the talk - time battery life of a group of cell phones is normally distributed with a mean of 5 hours and…

Explanation:

The talk-time battery life $X$ follows a normal distribution with mean $\mu = 5$ hours and standard deviation $\sigma = 15$ minutes. First, convert the standard deviation to hours: $\sigma = \frac{15}{60} = 0.25$ hours. So, $X \sim N(5, 0.25^2)$. We will use the standard normal distribution $Z = \frac{X - \mu}{\sigma}$ to find the required percentages.

a) Percentage of phones with battery life of at least 4 hours and 45 minutes

Step1: Convert the time to hours.

4 hours and 45 minutes is $4 + \frac{45}{60} = 4.75$ hours. We need to find $P(X \ge 4.75)$.

Step2: Calculate the z-score.

The z-score corresponding to $X = 4.75$ is $Z = \frac{4.75 - \mu}{\sigma}$. $$ Z = \frac{4.75 - 5}{0.25} = \frac{-0.25}{0.25} = -1 $$

Step3: Find the probability using the standard normal distribution.

We need to find $P(Z \ge -1)$. Using the symmetry of the normal distribution, $P(Z \ge -1) = P(Z \le 1)$. Using a standard normal table or calculator, $P(Z \le 1) \approx 0.8413$.

Step4: Convert the probability to a percentage.

The percentage is $0.8413 \times 100% = 84.13%$.

b) Percentage of phones with battery life between 4.5 hours and 5.25 hours

Step1: Identify the range in hours.

We need to find the probability $P(4.5 \le X \le 5.25)$.

Step2: Calculate the z-scores for the boundaries.

For $X_1 = 4.5$: $$ Z_1 = \frac{4.5 - 5}{0.25} = \frac{-0.5}{0.25} = -2 $$ For $X_2 = 5.25$: $$ Z_2 = \frac{5.25 - 5}{0.25} = \frac{0.25}{0.25} = 1 $$

Step3: Find the probability using the standard normal distribution.

We need to find $P(-2 \le Z \le 1)$. This is calculated as $P(Z \le 1) - P(Z < -2)$. Using a standard normal table or calculator: $P(Z \le 1) \approx 0.8413$. $P(Z < -2) \approx 0.0228$. $P(-2 \le Z \le 1) = P(Z \le 1) - P(Z < -2) \approx 0.8413 - 0.0228 = 0.8185$.

Step4: Convert the probability to a percentage.

The percentage is $0.8185 \times 100% = 81.85%$.

c) Percentage of phones with battery life less than 5 hours or greater than 5.5 hours

Step1: Identify the required probabilities.

We need to find $P(X < 5 \text{ or } X > 5.5)$, which is $P(X < 5) + P(X > 5.5)$.

Step2: Calculate the z-scores for the boundaries.

For $X_1 = 5$: $$ Z_1 = \frac{5 - 5}{0.25} = 0 $$ For $X_2 = 5.5$: $$ Z_2 = \frac{5.5 - 5}{0.25} = \frac{0.5}{0.25} = 2 $$

Step3: Find the probability using the standard normal distribution.

We need to find $P(Z < 0) + P(Z > 2)$. $P(Z < 0) = 0.5$ (due to symmetry of the normal distribution around the mean). $P(Z > 2) = 1 - P(Z \le 2)$. Using a standard normal table or calculator, $P(Z \le 2) \approx 0.9772$. $P(Z > 2) = 1 - 0.9772 = 0.0228$. The total probability is $P(Z < 0) + P(Z > 2) = 0.5 + 0.0228 = 0.5228$.

Step4: Convert the probability to a percentage.

The percentage is $0.5228 \times 100% = 52.28%$.

Answer:

a) 84.13% b) 81.85% c) 52.28%