In a recent survey of 683 working Americans ages 25-34, the average weekly amount spent on lunch as 46.18 with standard deviation 2.74. The weekly amounts are approximately bell-shaped.

Question

(a) Estimate the percentage of amounts that were less than 40.70. Round the answer to one decimal place.

Answer 1

To estimate the percentage of amounts that were less than $40.70, we can use the properties of the normal distribution since the weekly amounts are described as being approximately bell-shaped (normally distributed).

1. Calculate the z-score for the value of $40.70 using the formula:

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

where \( X \) is the value we are interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Given:

• \( \mu = 46.18 \)
• \( \sigma = 2.74 \)
• \( X = 40.70 \)

Plugging in the values:

\[ z = \frac{40.70 - 46.18}{2.74} = \frac{-5.48}{2.74} \approx -2.00 \]

2. Look up the z-score in the standard normal distribution table, or use a calculator that provides cumulative probabilities for the normal distribution.

A z-score of -2.00 corresponds to approximately 0.0228, or 2.28%. This indicates that about 2.28% of the amounts spent on lunch are less than $40.70.

3. Round to one decimal place:

Thus, the estimated percentage of amounts that were less than $40.70 is approximately:

\[ \boxed{2.3%} \]

Answer 2

estimate the percentage of amounts that were greater than 48.92. Round the answer to one decimal place.

Answer 3

To estimate the percentage of amounts that were greater than $48.92, we will once again use the z-score formula for the normal distribution.

1. Calculate the z-score for the value of $48.92:

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

where \( X \) is the value we are interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Given:

• \( \mu = 46.18 \)
• \( \sigma = 2.74 \)
• \( X = 48.92 \)

Plugging in the values:

\[ z = \frac{48.92 - 46.18}{2.74} = \frac{2.74}{2.74} = 1.00 \]

2. Look up the z-score of 1.00 in the standard normal distribution table, or use a calculator.

The cumulative probability for a z-score of 1.00 is approximately 0.8413. This means that about 84.13% of the amounts are less than $48.92.

3. Calculate the percentage greater than $48.92:

To find the percentage of amounts greater than $48.92, we can subtract the cumulative probability from 1:

\[ P(X > 48.92) = 1 - P(X < 48.92) = 1 - 0.8413 = 0.1587 \]

4. Convert to percentage and round:

\[ 0.1587 \times 100 \approx 15.87% \]

Rounding to one decimal place, the estimated percentage of amounts that were greater than $48.92 is approximately:

\[ \boxed{15.9%} \]

Answer 4

(c) Estimate the percentage of amounts that were between 46.18 and 51.66. Round the answer to one decimal place.

Answer 5

To estimate the percentage of amounts that were between $46.18 and $51.66, we will calculate the z-scores for both values and find the corresponding cumulative probabilities.

1. Calculate the z-score for $46.18:

Since $46.18 is the mean (\( \mu \)), its z-score will be:

\[ z = \frac{46.18 - 46.18}{2.74} = 0 \]

2. Calculate the z-score for $51.66:

Using the z-score formula:

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

where \( X = 51.66 \), \( \mu = 46.18 \), and \( \sigma = 2.74 \):

\[ z = \frac{51.66 - 46.18}{2.74} = \frac{5.48}{2.74} \approx 2.00 \]

3. Look up the cumulative probabilities:

• For \( z = 0 \), the cumulative probability is 0.5000 (since it's the mean).
• For \( z = 2.00 \), the cumulative probability is approximately 0.9772.

4. Calculate the percentage of amounts between $46.18 and $51.66:

We find the difference in cumulative probabilities:

\[ P(46.18 < X < 51.66) = P(X < 51.66) - P(X < 46.18) \] \[ P(46.18 < X < 51.66) = 0.9772 - 0.5000 = 0.4772 \]

5. Convert to percentage and round:

\[ 0.4772 \times 100 \approx 47.72% \]

Rounding to one decimal place, the estimated percentage of amounts that were between $46.18 and $51.66 is approximately:

\[ \boxed{47.7%} \]