Questions: Lunch break: In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was 44.54 with standard deviation 3.09. The weekly amounts are approximately bell-shaped. Part: 0 / 3 Part 1 of 3 (a) Estimate the percentage of amounts that are between 41.45 and 47.63. (Choose one) of the amounts fall between 41.45 and 47.63. Approximately 68% Approximately 75% Approximately 95% Approximately 88.9% Almost all

Solution Steps

To estimate the percentage of amounts that fall between \$41.45 and \$47.63, we can use the empirical rule (68-95-99.7 rule) for a normal distribution. This rule states that approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. We will calculate how many standard deviations away the values \$41.45 and \$47.63 are from the mean and use this information to estimate the percentage.

To determine how many standard deviations the bounds are from the mean, we calculate the z-scores for the lower and upper bounds using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where \( X \) is the bound, \( \mu = 44.54 \) is the mean, and \( \sigma = 3.09 \) is the standard deviation.

For the lower bound (\$41.45):

\[ z_{\text{lower}} = \frac{41.45 - 44.54}{3.09} \approx -1.0000 \]

For the upper bound (\$47.63):

\[ z_{\text{upper}} = \frac{47.63 - 44.54}{3.09} \approx 1.0000 \]

Using the z-scores, we find the cumulative probabilities from the standard normal distribution.

For \( z_{\text{lower}} \approx -1.0000 \):

\[ P(Z < -1.0000) \approx 0.1587 \]

For \( z_{\text{upper}} \approx 1.0000 \):

\[ P(Z < 1.0000) \approx 0.8413 \]

The percentage of amounts between the bounds is the difference between the cumulative probabilities:

\[ P(41.45 < X < 47.63) = P(Z < 1.0000) - P(Z < -1.0000) \approx 0.8413 - 0.1587 = 0.6826 \]

Convert this probability to a percentage:

\[ \text{Percentage} = 0.6826 \times 100\% = 68.26\% \]

Final Answer