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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.

7 answers

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.

this is incorrect

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!

I apologize, you were indeed correct.

b) Estimate the percentage of amounts that are between 36.63 and 54.15.

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.

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

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.