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.
(a) Estimate the percentage of amounts that were less than 40.70. Round the answer to one decimal place.
Answers
Answer
estimate the percentage of amounts that were greater than 48.92. Round the answer to one decimal place.
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 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:
\[
\boxed{2.3\%}
\]
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:
\[
\boxed{2.3\%}
\]
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:
\[
\boxed{15.9\%}
\]
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:
\[
\boxed{15.9\%}
\]
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:
\[
\boxed{47.7\%}
\]
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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