Question

In a recent survey of 683 working Americans ages 25-34, the average weekly amount spent on lunch as 46.18 with standard deviation 2.74. The weekly amounts are approximately bell-shaped.

(a) Estimate the percentage of amounts that were less than 40.70. Round the answer to one decimal place.

Answers

Answered by GPT-4o mini
To estimate the percentage of amounts that were less than $40.70, we can use the properties of the normal distribution since the weekly amounts are described as being approximately bell-shaped (normally distributed).

1. **Calculate the z-score** for the value of $40.70 using the formula:

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

where \( X \) is the value we are interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Given:
- \( \mu = 46.18 \)
- \( \sigma = 2.74 \)
- \( X = 40.70 \)

Plugging in the values:

\[
z = \frac{40.70 - 46.18}{2.74} = \frac{-5.48}{2.74} \approx -2.00
\]

2. **Look up the z-score** in the standard normal distribution table, or use a calculator that provides cumulative probabilities for the normal distribution.

A z-score of -2.00 corresponds to approximately 0.0228, or 2.28%. This indicates that about 2.28% of the amounts spent on lunch are less than $40.70.

3. **Round to one decimal place**:

Thus, the estimated percentage of amounts that were less than $40.70 is approximately:

\[
\boxed{2.3\%}
\]
Answer
estimate the percentage of amounts that were greater than 48.92. Round the answer to one decimal place.
Answered by GPT-4o mini
To estimate the percentage of amounts that were greater than $48.92, we will once again use the z-score formula for the normal distribution.

1. **Calculate the z-score** for the value of $48.92:

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

where \( X \) is the value we are interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Given:

- \( \mu = 46.18 \)
- \( \sigma = 2.74 \)
- \( X = 48.92 \)

Plugging in the values:

\[
z = \frac{48.92 - 46.18}{2.74} = \frac{2.74}{2.74} = 1.00
\]

2. **Look up the z-score** of 1.00 in the standard normal distribution table, or use a calculator.

The cumulative probability for a z-score of 1.00 is approximately 0.8413. This means that about 84.13% of the amounts are less than $48.92.

3. **Calculate the percentage greater than $48.92**:

To find the percentage of amounts greater than $48.92, we can subtract the cumulative probability from 1:

\[
P(X > 48.92) = 1 - P(X < 48.92) = 1 - 0.8413 = 0.1587
\]

4. **Convert to percentage** and round:

\[
0.1587 \times 100 \approx 15.87\%
\]

Rounding to one decimal place, the estimated percentage of amounts that were greater than $48.92 is approximately:

\[
\boxed{15.9\%}
\]
Answer
(c) Estimate the percentage of amounts that were between 46.18 and 51.66. Round the answer to one decimal place.
Answered by GPT-4o mini
To estimate the percentage of amounts that were between $46.18 and $51.66, we will calculate the z-scores for both values and find the corresponding cumulative probabilities.

1. **Calculate the z-score** for $46.18:

Since $46.18 is the mean (\( \mu \)), its z-score will be:

\[
z = \frac{46.18 - 46.18}{2.74} = 0
\]

2. **Calculate the z-score** for $51.66:

Using the z-score formula:

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

where \( X = 51.66 \), \( \mu = 46.18 \), and \( \sigma = 2.74 \):

\[
z = \frac{51.66 - 46.18}{2.74} = \frac{5.48}{2.74} \approx 2.00
\]

3. **Look up the cumulative probabilities**:

- For \( z = 0 \), the cumulative probability is 0.5000 (since it's the mean).
- For \( z = 2.00 \), the cumulative probability is approximately 0.9772.

4. **Calculate the percentage of amounts between $46.18 and $51.66**:

We find the difference in cumulative probabilities:

\[
P(46.18 < X < 51.66) = P(X < 51.66) - P(X < 46.18)
\]
\[
P(46.18 < X < 51.66) = 0.9772 - 0.5000 = 0.4772
\]

5. **Convert to percentage** and round:

\[
0.4772 \times 100 \approx 47.72\%
\]

Rounding to one decimal place, the estimated percentage of amounts that were between $46.18 and $51.66 is approximately:

\[
\boxed{47.7\%}
\]

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