Problem 21 - Algebra 2 and Trigonometry

Question

When adjusted for inflation, the monthly amount that a historic restaurant spent on cleaning for a 30 -year period was normally distributed with a mean of \(\$ 2,100\) and a standard deviation of \(\$ 240 .\) a. What is the probability that the restaurant will spend between \(\$ 2,500\) and \(\$ 2,800\) on cleaning for the month of January? b. If the restaurant has only \(\$ 2,400\) allotted for cleaning for the month of January, what is the probability that it will exceed its budget for cleaning?

Step-by-Step Solution

Answer

a. 0.0457 b. 0.1056

Step 1: Understand the problem context

We are given a normal distribution for the monthly spending on cleaning at a historic restaurant. The mean is \( \\(2100 \) and the standard deviation is \( \\)240 \). We need to calculate probabilities related to this distribution.

Step 2: Standardizing the distribution

To find probabilities, we'll convert the spending amounts to \( Z \)-scores using the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Step 3: Calculate Z-scores for part (a)

We need to find the probability that the spending is between \( \\(2,500 \) and \( \\)2,800 \). Calculate the \( Z \)-scores for these values:- For \( \\(2,500 \), \( Z = \frac{2500 - 2100}{240} = \frac{400}{240} \approx 1.67 \).- For \( \\)2,800 \), \( Z = \frac{2800 - 2100}{240} = \frac{700}{240} \approx 2.92 \).

Step 4: Find probabilities using Z-scores for part (a)

Use the standard normal distribution table to find the probabilities:- Probability for \( Z=1.67 \) is approximately \( 0.9525 \).- Probability for \( Z=2.92 \) is approximately \( 0.9982 \).Calculate the probability for spending between \( \\(2,500 \) and \( \\)2,800 \): \( P(1.67 < Z < 2.92) = 0.9982 - 0.9525 = 0.0457 \).

Step 5: Calculate Z-score for part (b)

For part (b), we need the probability that spending exceeds \( \\(2,400 \). Calculate the \( Z \)-score for \( \\)2,400 \):- \( Z = \frac{2400 - 2100}{240} = \frac{300}{240} = 1.25 \).

Step 6: Find probability using Z-score for part (b)

Use the standard normal distribution table to find probabilities:- Probability for \( Z=1.25 \) is approximately \( 0.8944 \).The probability that spending exceeds \( \$2,400 \) is \( P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056 \).

Key Concepts

Understanding Standard Deviation

Standard deviation is a measure of how spread out the values in a data set are. In simpler terms, it gives us an idea of how much individual data points differ from the mean, or average, value.

For example, in the context of the restaurant's cleaning costs, if the average amount spent is \( \\(2,100 \) with a standard deviation of \( \\)240 \), this means that most of the monthly cleaning costs are likely within \( \\(240 \) above or below the average.

• If costs are consistently very close to \( \\)2,100 \), then the standard deviation is smaller, indicating less variability.
• If costs vary greatly, the standard deviation is larger. This shows there is more variability, or spread, in the monthly spending amounts.

Understanding the standard deviation allows businesses to predict and budget more accurately.

Explaining Z-scores

A Z-score tells us how many standard deviations a specific value is from the mean of a distribution. It provides a way to standardize different data points for comparison.

In the exercise, if we want to know about spending amounts \( \\(2,500 \) and \( \\)2,800 \):

• Calculate the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the specific value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
• For example, a \( Z \)-score of 1.67 for \( \$2,500 \) shows it's 1.67 standard deviations above the mean.
• Higher or lower Z-scores indicate more extreme values compared to the overall distribution.

Z-scores help in comparing individual values against the entire set.

Concept of the Standard Normal Distribution

The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. This allows us to directly use Z-scores and standard normal distribution tables to find probabilities.

Here’s why it’s useful:

• Once we convert our data to Z-scores, we can use the standard normal distribution to find probabilities for different ranges.
• For example, to find the probability that spending falls between certain values, we convert the spending amounts to Z-scores and then use those Z-scores with a standard normal distribution table to find our desired probabilities.

It simplifies the process of calculating probabilities across varying contexts.

Probability Calculation in the Context of Normal Distribution

Calculating probabilities in a normal distribution involves understanding where a data point falls relative to the entire distribution.

For the restaurant spending:

• After converting spending amounts into Z-scores, use the standard normal distribution table to find the probabilities associated with those Z-scores.
• For the range \( \\(2,500 \) to \( \\)2,800 \), you calculate \[ P(1.67 < Z < 2.92) = 0.9982 - 0.9525 = 0.0457 \] This represents the probability of spending between these amounts.
• For exceeding the budget of \( \\(2,400 \), calculate \[ P(Z > 1.25) = 1 - P(Z < 1.25) = 0.1056 \] This probability indicates the chance that the spending will exceed \( \\)2,400 \).

Such probability calculations help in budgeting and forecasting under uncertainty.