Question

Please make sure to properly format your answers. All dollars figures in the answers need to include the dollar sign and any amount over 1,000 should include the comma ($2,354.67). All percentage values in the answers need to include a percentage (%) sign. For all items without specific rounding instructions, round your answers to two decimal places, show both decimal places (5.06).

A recent survey by the American Automobile Association showed that a family of two adults and two
children on vacation in the United States will pay an average of $247 per day for food and lodging with a
standard deviation of $60 per day. Assuming the data are normally distributed, if the family had a z-score
of 1.5, what was their daily expense for food and lodging?
If the data are normally distributed, find the percentage of these families who spent:

1. Less than $167 per day.

2.Less than $367 per day.

3.More than $247 per day.

4.More than $350 per day.

5.Less than $67 per day.

6.Between $200 and $300 per day.

7.Between $360 and $400 per day.

8.More than the median.

9.Less than the mean.

Answer 1

Explanation:

To find the daily expense for food and lodging for the family, we can use the formula:

z = (x - μ) / σ

where z is the z-score, x is the daily expense for food and lodging, μ is the mean daily expense, and σ is the standard deviation. Rearranging this formula to solve for x, we get:

x = z * σ + μ

Substituting z = 1.5, σ = $60, and μ = $247, we get:

x = 1.5 * $60 + $247 = $337

Therefore, the daily expense for food and lodging for the family was $337.

To find the percentage of families who spent less than $367 per day, we can standardize the value using the same formula as above:

z = (x - μ) / σ

Substituting x = $367, μ = $247, and σ = $60, we get:

z = ($367 - $247) / $60 = 2

Using a standard normal distribution table or calculator, we can find the percentage of families with a z-score less than 2, which is approximately 97.72%. Therefore, approximately 97.72% of families spent less than $367 per day.

To find the percentage of families who spent more than $247 per day, we can use the same formula and standardize the value:

z = (x - μ) / σ

Substituting x = $247, μ = $247, and σ = $60, we get:

z = ($247 - $247) / $60 = 0

Using a standard normal distribution table or calculator, we can find the percentage of families with a z-score greater than 0, which is 50%. Therefore, 50% of families spent more than $247 per day.

To find the percentage of families who spent more than $350 per day, we can standardize the value:

z = (x - μ) / σ

Substituting x = $350, μ = $247, and σ = $60, we get:

z = ($350 - $247) / $60 = 1.72

Using a standard normal distribution table or calculator, we can find the percentage of families with a z-score greater than 1.72, which is approximately 4.35%. Therefore, approximately 4.35% of families spent more than $350 per day.

To find the percentage of families who spent less than $67 per day, we can standardize the value:

z = (x - μ) / σ

Substituting x = $67, μ = $247, and σ = $60, we get:

z = ($67 - $247) / $60 = -3.00

Using a standard normal distribution table or calculator, we can find the percentage of families with a z-score less than -3.00, which is approximately 0.13%. Therefore, approximately 0.13% of families spent less than $67 per day.

To find the percentage of families who spent between $200 and $300 per day, we need to standardize both values and find the area between them. Using the same formula as above:

z1 = ($200 - $247) / $60 = -0.78

z2 = ($300 - $247) / $60 = 0.78

Using a standard normal distribution table or calculator, we can find the percentage of families with a z-score between -0.78 and 0.78, which is approximately 53.91%. Therefore, approximately 53.91% of families spent between