Question

In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was 45.39 with standard deviation 2.92. The weekly amounts are approximately bell-shaped.
a) Estimate the percentage of amounts that are between 39.55 and 51.23.
Answer
this is incorrect
Answer
I apologize, you were indeed correct.

b) Estimate the percentage of amounts that are between 36.63 and 54.15.
Answer
c) Between what two values will approximately 68% of the amounts be?
Answered by GPT-4o mini
To estimate the percentage of amounts spent on lunch that fall between $39.55 and $51.23, we can use the properties of the normal (bell-shaped) distribution.

Given:
- Mean (\( \mu \)) = 45.39
- Standard Deviation (\( \sigma \)) = 2.92

First, we need to calculate the z-scores for the values 39.55 and 51.23.

The z-score is calculated using the formula:

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

### Calculate the z-score for 39.55:
\[
z_{39.55} = \frac{(39.55 - 45.39)}{2.92} = \frac{-5.84}{2.92} \approx -2.00
\]

### Calculate the z-score for 51.23:
\[
z_{51.23} = \frac{(51.23 - 45.39)}{2.92} = \frac{5.84}{2.92} \approx 2.00
\]

### Look up the z-scores in the standard normal distribution table:
A z-score of -2.00 corresponds to a cumulative probability of approximately 0.0228 (or 2.28%), and a z-score of 2.00 corresponds to approximately 0.9772 (or 97.72%).

### Calculate the percentage of amounts between 39.55 and 51.23:
To find the percentage of amounts that fall between these two values, we subtract the cumulative probability of the lower z-score from that of the upper z-score:

\[
P(39.55 < X < 51.23) = P(Z < 2.00) - P(Z < -2.00) \approx 0.9772 - 0.0228 = 0.9544
\]

### Convert to percentage:
\[
0.9544 \times 100 \approx 95.44\%
\]

So, approximately **95.44%** of the amounts spent on lunch are between $39.55 and $51.23.
Answered by GPT-4o mini
I apologize for the oversight. Let's go through the calculations again step by step.

### Given Data
- Mean (\(\mu\)) = 45.39
- Standard Deviation (\(\sigma\)) = 2.92

### Step 1: Calculate the z-scores

**For \(X = 39.55\)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(39.55 - 45.39)}{2.92} = \frac{-5.84}{2.92} \approx -2.00
\]

**For \(X = 51.23\)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(51.23 - 45.39)}{2.92} = \frac{5.84}{2.92} \approx 2.00
\]

### Step 2: Look up the z-scores

Using the standard normal distribution table or calculator:

- **For \(z = -2.00\)**:
- Cumulative probability \(P(Z < -2.00) \approx 0.0228\)

- **For \(z = 2.00\)**:
- Cumulative probability \(P(Z < 2.00) \approx 0.9772\)

### Step 3: Calculate the percentage between the two z-scores

Now we need to find the probability that \(X\) falls between these two z-scores:

\[
P(39.55 < X < 51.23) = P(Z < 2.00) - P(Z < -2.00)
\]
\[
P(39.55 < X < 51.23) = 0.9772 - 0.0228 = 0.9544
\]

### Step 4: Convert to Percentage

Convert the probability to a percentage:

\[
0.9544 \times 100 \approx 95.44\%
\]

### Conclusion

After verifying the calculations, the estimate is indeed correct, and approximately **95.44%** of the amounts that working Americans ages 25-34 spend on lunch fall between $39.55 and $51.23.

If there's a specific aspect of the problem you'd like to discuss or if there was an expected different aspect to the answer, please let me know!
Answered by GPT-4o mini
No problem at all! Let's estimate the percentage of amounts spent on lunch that fall between $36.63 and $54.15 using the same mean and standard deviation provided earlier.

### Given Data
- Mean (\(\mu\)) = 45.39
- Standard Deviation (\(\sigma\)) = 2.92

### Step 1: Calculate the z-scores

**For \(X = 36.63\)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(36.63 - 45.39)}{2.92} = \frac{-8.76}{2.92} \approx -3.00
\]

**For \(X = 54.15\)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(54.15 - 45.39)}{2.92} = \frac{8.76}{2.92} \approx 3.00
\]

### Step 2: Look up the z-scores

Using the standard normal distribution table or calculator:

- **For \(z = -3.00\)**:
- Cumulative probability \(P(Z < -3.00) \approx 0.0013\) (or 0.13%)

- **For \(z = 3.00\)**:
- Cumulative probability \(P(Z < 3.00) \approx 0.9987\) (or 99.87%)

### Step 3: Calculate the percentage between the two z-scores

Now, calculate the probability that \(X\) falls between these two z-scores:

\[
P(36.63 < X < 54.15) = P(Z < 3.00) - P(Z < -3.00)
\]
\[
P(36.63 < X < 54.15) = 0.9987 - 0.0013 = 0.9974
\]

### Step 4: Convert to Percentage

Convert the probability to a percentage:

\[
0.9974 \times 100 \approx 99.74\%
\]

### Conclusion

Thus, approximately **99.74%** of the amounts that working Americans ages 25-34 spend on lunch fall between $36.63 and $54.15.
Answered by GPT-4o mini
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. Given the mean and standard deviation from the previous data:

- Mean (\(\mu\)) = 45.39
- Standard Deviation (\(\sigma\)) = 2.92

### Step 1: Calculate the range

To find the values that encompass approximately 68% of the amounts, we calculate:

- Lower limit: \( \mu - \sigma \)
- Upper limit: \( \mu + \sigma \)

### Step 2: Perform the calculations

#### Lower limit:
\[
\text{Lower limit} = \mu - \sigma = 45.39 - 2.92 \approx 42.47
\]

#### Upper limit:
\[
\text{Upper limit} = \mu + \sigma = 45.39 + 2.92 \approx 48.31
\]

### Conclusion

Therefore, approximately 68% of the amounts spent on lunch by working Americans ages 25-34 will be between **$42.47** and **$48.31**.