Question

In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was 46.43 with standard deviation 2.90. The weekly amounts are approximately bell-shaped.

Estimate the percentage of amounts that are between 40.63 and 52.23.

Answer 1

(b) Estimate the percentage of amounts that are between 37.73 and 55.13

Answer 2

(c) Between what two values will approximately 68% of the amounts be?

Answer 3

To estimate the percentage of amounts spent on lunch that fall between $40.63 and $52.23, we can use the properties of a bell-shaped (normal) distribution, specifically the empirical rule (68-95-99.7 rule).

Given:
- Mean (\(\mu\)) = $46.43
- Standard deviation (\(\sigma\)) = $2.90

First, we can calculate how many standard deviations the values of $40.63 and $52.23 are from the mean:

1. Calculate the z-score for $40.63:

\[
z_1 = \frac{X - \mu}{\sigma} = \frac{40.63 - 46.43}{2.90} = \frac{-5.80}{2.90} \approx -2.00
\]

2. Calculate the z-score for $52.23:

\[
z_2 = \frac{X - \mu}{\sigma} = \frac{52.23 - 46.43}{2.90} = \frac{5.80}{2.90} \approx 2.00
\]

Now, we have:
- \( z_1 \approx -2 \)
- \( z_2 \approx 2 \)

According to the empirical rule:
- Approximately 95% of the data lies within 2 standard deviations from the mean in a normal distribution.

Thus, the percentage of amounts spent on lunch that are between $40.63 and $52.23 is approximately **95%**.

Answer 4

To estimate the percentage of amounts spent on lunch that fall between $37.73 and $55.13, we will again use the properties of a normal distribution.

Given:
- Mean (\(\mu\)) = $46.43
- Standard deviation (\(\sigma\)) = $2.90

We can calculate the z-scores for both $37.73 and $55.13.

1. Calculate the z-score for $37.73:

\[
z_1 = \frac{X - \mu}{\sigma} = \frac{37.73 - 46.43}{2.90} = \frac{-8.70}{2.90} \approx -3.00
\]

2. Calculate the z-score for $55.13:

\[
z_2 = \frac{X - \mu}{\sigma} = \frac{55.13 - 46.43}{2.90} = \frac{8.70}{2.90} \approx 3.00
\]

Now, we have:
- \( z_1 \approx -3 \)
- \( z_2 \approx 3 \)

According to the empirical rule:
- Approximately 99.7% of the data falls within 3 standard deviations from the mean in a normal distribution.

Therefore, the percentage of amounts spent on lunch that are between $37.73 and $55.13 is approximately **99.7%**.

Answer 5

To find the two values between which approximately 68% of the amounts will fall in a normal distribution, we can use the empirical rule (68-95-99.7 rule), which states that about 68% of the data lies within 1 standard deviation from the mean.

Given:
- Mean (\(\mu\)) = $46.43
- Standard Deviation (\(\sigma\)) = $2.90

To find the range that contains approximately 68% of the amounts, we calculate:

1. Lower Bound:
\[
\text{Lower Bound} = \mu - \sigma = 46.43 - 2.90 \approx 43.53
\]

2. Upper Bound:
\[
\text{Upper Bound} = \mu + \sigma = 46.43 + 2.90 \approx 49.33
\]

Thus, approximately 68% of the amounts spent on lunch will fall between **$43.53** and **$49.33**.