Question
One year consumers spent an average of $24 on a meal at a resturant. Assume that the amount spent on a resturant meal is normally distributed and that the standard deviation is $4. Complete parts (a) through (c) below.
Answers
GPT-4o mini
To analyze the situation described regarding the spending on a restaurant meal, let's break it down into parts (a), (b), and (c). Since the specifics of parts (a), (b), and (c) are not provided, I’ll outline typical questions that could be asked in this context and how you might approach them.
### (a) Find the probability that a randomly selected consumer spends less than $20 on a meal.
To find this probability, we can use the Z-score formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \( X \) is the value we are interested in ($20 in this case).
- \( \mu = 24 \) (the mean).
- \( \sigma = 4 \) (the standard deviation).
First, we calculate the Z-score for $20:
\[
Z = \frac{(20 - 24)}{4} = \frac{-4}{4} = -1
\]
Now, we can use the standard normal distribution table (or a calculator) to find the probability corresponding to \( Z = -1 \).
Looking up \( Z = -1 \), we find that the probability \( P(Z < -1) \approx 0.1587 \).
### Conclusion for (a):
The probability that a randomly selected consumer spends less than $20 on a meal is approximately **0.1587** or **15.87%**.
---
### (b) Find the probability that a randomly selected consumer spends between $20 and $28 on a meal.
To find this probability, we will calculate the Z-scores for both $20 and $28.
For \( X = 20 \):
\[
Z_{20} = \frac{(20 - 24)}{4} = -1
\]
For \( X = 28 \):
\[
Z_{28} = \frac{(28 - 24)}{4} = 1
\]
Now using the standard normal distribution table:
- \( P(Z < 1) \approx 0.8413 \)
- \( P(Z < -1) \approx 0.1587 \)
Now, to find the probability between $20 and $28:
\[
P(20 < X < 28) = P(Z < 1) - P(Z < -1) \approx 0.8413 - 0.1587 = 0.6826
\]
### Conclusion for (b):
The probability that a randomly selected consumer spends between $20 and $28 on a meal is approximately **0.6826** or **68.26%**.
---
### (c) Find the amount spent such that 90% of consumers spend less than this amount.
To find this amount, we look for a Z-score that corresponds to 90% in the standard normal distribution table. The Z-score for 90% is approximately **1.2816**.
Now, we will use the Z-score formula to find the corresponding spending amount \( X \):
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Rearranging this we have:
\[
X = Z \cdot \sigma + \mu
\]
Substituting the values we found:
\[
X = (1.2816)(4) + 24 \approx 5.1264 + 24 \approx 29.1264
\]
### Conclusion for (c):
The amount spent such that 90% of consumers spend less than this amount is approximately **$29.13**.
---
In a real-world setting, you would use a normal distribution calculator or software to perform these calculations more accurately if needed. However, the above provides a good theoretical grounding.
### (a) Find the probability that a randomly selected consumer spends less than $20 on a meal.
To find this probability, we can use the Z-score formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \( X \) is the value we are interested in ($20 in this case).
- \( \mu = 24 \) (the mean).
- \( \sigma = 4 \) (the standard deviation).
First, we calculate the Z-score for $20:
\[
Z = \frac{(20 - 24)}{4} = \frac{-4}{4} = -1
\]
Now, we can use the standard normal distribution table (or a calculator) to find the probability corresponding to \( Z = -1 \).
Looking up \( Z = -1 \), we find that the probability \( P(Z < -1) \approx 0.1587 \).
### Conclusion for (a):
The probability that a randomly selected consumer spends less than $20 on a meal is approximately **0.1587** or **15.87%**.
---
### (b) Find the probability that a randomly selected consumer spends between $20 and $28 on a meal.
To find this probability, we will calculate the Z-scores for both $20 and $28.
For \( X = 20 \):
\[
Z_{20} = \frac{(20 - 24)}{4} = -1
\]
For \( X = 28 \):
\[
Z_{28} = \frac{(28 - 24)}{4} = 1
\]
Now using the standard normal distribution table:
- \( P(Z < 1) \approx 0.8413 \)
- \( P(Z < -1) \approx 0.1587 \)
Now, to find the probability between $20 and $28:
\[
P(20 < X < 28) = P(Z < 1) - P(Z < -1) \approx 0.8413 - 0.1587 = 0.6826
\]
### Conclusion for (b):
The probability that a randomly selected consumer spends between $20 and $28 on a meal is approximately **0.6826** or **68.26%**.
---
### (c) Find the amount spent such that 90% of consumers spend less than this amount.
To find this amount, we look for a Z-score that corresponds to 90% in the standard normal distribution table. The Z-score for 90% is approximately **1.2816**.
Now, we will use the Z-score formula to find the corresponding spending amount \( X \):
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Rearranging this we have:
\[
X = Z \cdot \sigma + \mu
\]
Substituting the values we found:
\[
X = (1.2816)(4) + 24 \approx 5.1264 + 24 \approx 29.1264
\]
### Conclusion for (c):
The amount spent such that 90% of consumers spend less than this amount is approximately **$29.13**.
---
In a real-world setting, you would use a normal distribution calculator or software to perform these calculations more accurately if needed. However, the above provides a good theoretical grounding.