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.
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Part 1
a. What is the probability that a randomly selected person spent more than ​$26​?

Answer 1

To find the probability that a randomly selected person spent more than $26 at a restaurant, we will use the properties of the normal distribution.

1. **Identify the parameters**:
- Mean (\(\mu\)): $22
- Standard deviation (\(\sigma\)): $5

2. **Calculate the Z-score** for $26:
\[
Z = \frac{X - \mu}{\sigma} = \frac{26 - 22}{5} = \frac{4}{5} = 0.8
\]

3. **Look up the Z-score** in the standard normal distribution table to find the probability of Z being less than 0.8.

The cumulative probability (area to the left of Z = 0.8) is approximately 0.7881.

4. **Calculate the probability of spending more than $26**:
The area to the right (which gives the probability of spending more than $26) is:
\[
P(X > 26) = 1 - P(Z < 0.8) = 1 - 0.7881 = 0.2119
\]

Thus, the probability that a randomly selected person spent more than $26 is approximately **0.2119**, or **21.19%**.