In a recent survey of 636 working Americans ages 25-34, the average weekly amount spent on lunch was $44.26 with standard deviation $2.68. The weekly amounts are approximately bell-shaped.

Question

a) Estimate the percentage of amounts that were less than $48.90. Round the answer to one decimal place.

Answer 1

To estimate the percentage of amounts spent on lunch that are less than $48.90, we can use the properties of the normal distribution since the data is approximately bell-shaped.

We are given:

• Mean (μ) = $44.26
• Standard deviation (σ) = $2.68

We first calculate the z-score for $48.90 using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where:

• \(X\) is the value we are interested in ($48.90),
• \(\mu\) is the mean,
• \(\sigma\) is the standard deviation.

Substituting in the values: \[ z = \frac{(48.90 - 44.26)}{2.68} = \frac{4.64}{2.68} \approx 1.7299 \approx 1.73 \]

Now we will look up the z-score of 1.73 in the standard normal distribution table or use a calculator/normal distribution function to find the cumulative probability.

For a z-score of 1.73, the cumulative probability is approximately 0.9582, or 95.82%.

Therefore, the percentage of amounts that were less than $48.90 is approximately: \[ \text{Percentage} \approx 95.8% \]

Rounding to one decimal place, the final answer is: 95.8%.

Answer 2

The correct answer is 2.5% amounts were less than $38.90.
Knowing this,
(b) Estimate the percentage of amounts that were greater than $46.94. Round the answer to one decimal place.

Answer 3

To estimate the percentage of amounts spent on lunch that are greater than $46.94, we will follow a similar process as before. We will first calculate the z-score for $46.94.

Given:

• Mean (μ) = $44.26
• Standard deviation (σ) = $2.68

Using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \]

Substituting in the values: \[ z = \frac{(46.94 - 44.26)}{2.68} = \frac{2.68}{2.68} \approx 1.0 \]

Now we will look up the z-score of 1.0 in the standard normal distribution table or use a calculator/normal distribution function to find the cumulative probability.

For a z-score of 1.0, the cumulative probability is approximately 0.8413 (or 84.13%).

To find the percentage of amounts that are greater than $46.94, we subtract this cumulative probability from 1:

\[ P(X > 46.94) = 1 - P(X < 46.94) = 1 - 0.8413 \approx 0.1587 \]

To convert this to a percentage: \[ \text{Percentage} \approx 0.1587 \times 100 \approx 15.87% \]

Rounding to one decimal place, the final answer is: 15.9%.

Answer 4

That is correct.
c) Estimate the percentage of amounts that were between $44.26 and $49.62. Round the answer to one decimal place.

Answer 5

To estimate the percentage of amounts spent on lunch that are between $44.26 and $49.62, we need to calculate the z-scores for both amounts using the following formula:

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

Given:

• Mean (μ) = $44.26
• Standard deviation (σ) = $2.68

1. Calculate the z-score for $44.26: \[ z_1 = \frac{(44.26 - 44.26)}{2.68} = \frac{0}{2.68} = 0 \]

   The cumulative probability for a z-score of 0 is 0.5000 (or 50%).

2. Calculate the z-score for $49.62: \[ z_2 = \frac{(49.62 - 44.26)}{2.68} = \frac{5.36}{2.68} \approx 2.0 \]

   The cumulative probability for a z-score of 2.0 is approximately 0.9772 (or 97.72%).

3. Calculate the percentage of amounts between $44.26 and $49.62: \[ P(44.26 < X < 49.62) = P(X < 49.62) - P(X < 44.26) = 0.9772 - 0.5000 \] \[ P(44.26 < X < 49.62) \approx 0.4772 \]

   To convert this to a percentage: \[ \text{Percentage} \approx 0.4772 \times 100 \approx 47.72% \]

Rounding to one decimal place, the final answer is: 47.7%.