Question
One year consumers spent an average of $22 on a meal at a resturant. Assume that the amount spent on a resturant meal is normally distributed and that the standard deviation is $5. Complete parts (a) through (c) below.
Answers
GPT-4o mini
To address your request effectively, let's break it down into parts (a) through (c). For statistical analysis involving normally distributed data, we typically follow certain steps. Assuming you want to calculate probabilities, percentiles, or similar statistics based on your provided data, here’s how to proceed:
**Given:**
- Mean (\(\mu\)) = $22
- Standard Deviation (\(\sigma\)) = $5
### (a) Find the probability that a consumer spends less than $20 on a meal.
To find this, we first need to convert the amount into a Z-score, which is given by the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \(X\) = the value we are interested in ($20)
- \(\mu\) = mean ($22)
- \(\sigma\) = standard deviation ($5)
Plugging in the values:
\[
Z = \frac{(20 - 22)}{5} = \frac{-2}{5} = -0.4
\]
Now, we look up the Z-score of -0.4 in the standard normal distribution table (or use a calculator) to find the probability.
Using a Z-table, the probability for \(Z = -0.4\) is approximately 0.3446.
Therefore, the probability that a consumer spends less than $20 is **approximately 34.46%**.
### (b) Find the probability that a consumer spends more than $25 on a meal.
Again, we calculate the Z-score:
\[
Z = \frac{(25 - 22)}{5} = \frac{3}{5} = 0.6
\]
Next, we find the probability corresponding to \(Z = 0.6\) in the Z-table.
Using a Z-table, the probability for \(Z = 0.6\) is approximately 0.7257. However, we need the probability of spending **more than** $25, so we subtract this from 1:
\[
P(X > 25) = 1 - P(Z < 0.6) = 1 - 0.7257 = 0.2743
\]
Thus, the probability that a consumer spends more than $25 is **approximately 27.43%**.
### (c) Find the amount spent such that 90% of consumers spend less than this amount.
To find this, we need to find the Z-score that corresponds to the 90th percentile. Looking up the Z-table, we find that the Z-score that corresponds to 0.90 is approximately 1.28.
Now, we can convert the Z-score back to the actual spending amount using the following formula:
\[
X = \mu + Z \cdot \sigma
\]
Substituting in the known values:
\[
X = 22 + (1.28 \cdot 5) = 22 + 6.4 = 28.4
\]
Thus, the amount spent such that 90% of consumers spend less than this amount is **approximately $28.40**.
### Summary:
(a) Probability of spending less than $20: **34.46%**
(b) Probability of spending more than $25: **27.43%**
(c) Amount such that 90% spend less: **$28.40**
**Given:**
- Mean (\(\mu\)) = $22
- Standard Deviation (\(\sigma\)) = $5
### (a) Find the probability that a consumer spends less than $20 on a meal.
To find this, we first need to convert the amount into a Z-score, which is given by the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \(X\) = the value we are interested in ($20)
- \(\mu\) = mean ($22)
- \(\sigma\) = standard deviation ($5)
Plugging in the values:
\[
Z = \frac{(20 - 22)}{5} = \frac{-2}{5} = -0.4
\]
Now, we look up the Z-score of -0.4 in the standard normal distribution table (or use a calculator) to find the probability.
Using a Z-table, the probability for \(Z = -0.4\) is approximately 0.3446.
Therefore, the probability that a consumer spends less than $20 is **approximately 34.46%**.
### (b) Find the probability that a consumer spends more than $25 on a meal.
Again, we calculate the Z-score:
\[
Z = \frac{(25 - 22)}{5} = \frac{3}{5} = 0.6
\]
Next, we find the probability corresponding to \(Z = 0.6\) in the Z-table.
Using a Z-table, the probability for \(Z = 0.6\) is approximately 0.7257. However, we need the probability of spending **more than** $25, so we subtract this from 1:
\[
P(X > 25) = 1 - P(Z < 0.6) = 1 - 0.7257 = 0.2743
\]
Thus, the probability that a consumer spends more than $25 is **approximately 27.43%**.
### (c) Find the amount spent such that 90% of consumers spend less than this amount.
To find this, we need to find the Z-score that corresponds to the 90th percentile. Looking up the Z-table, we find that the Z-score that corresponds to 0.90 is approximately 1.28.
Now, we can convert the Z-score back to the actual spending amount using the following formula:
\[
X = \mu + Z \cdot \sigma
\]
Substituting in the known values:
\[
X = 22 + (1.28 \cdot 5) = 22 + 6.4 = 28.4
\]
Thus, the amount spent such that 90% of consumers spend less than this amount is **approximately $28.40**.
### Summary:
(a) Probability of spending less than $20: **34.46%**
(b) Probability of spending more than $25: **27.43%**
(c) Amount such that 90% spend less: **$28.40**