We are given that the amount of money spent per lunch patron follows a normal distribution:

$X \sim N(5.87, 2.04^2)$

Since we are selecting a sample of 9 patrons, the sample mean follows the distribution:

$\bar{X} \sim N(5.87, \frac{2.04}{\sqrt{9}}) = N(5.87, 0.68^2)$

1. Probability that a single patron’s lunch cost is between $5.29 and $5.96$

For a single patron ($X$), we standardize using the z-score formula:

$Z = \frac{X - \mu}{\sigma}$

For $X = 5.29$:

$Z = \frac{5.29 - 5.87}{2.04} = \frac{-0.58}{2.04} = -0.2843$

For $X = 5.96$:

$Z = \frac{5.96 - 5.87}{2.04} = \frac{0.09}{2.04} = 0.0441$

Using the standard normal table:

$P(Z < -0.2843) \approx 0.3882, \quad P(Z < 0.0441) \approx 0.5176$

$P(5.29 < X < 5.96) = P(-0.2843 < Z < 0.0441) = 0.5176 - 0.3882 = 0.1294$

2. Probability that the average lunch cost for 9 patrons is between $5.29 and $5.96$

For the sample mean ($\bar{X}$), we use:

$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$

For $\bar{X} = 5.29$:

$Z = \frac{5.29 - 5.87}{0.68} = \frac{-0.58}{0.68} = -0.8529$

For $\bar{X} = 5.96$:

$Z = \frac{5.96 - 5.87}{0.68} = \frac{0.09}{0.68} = 0.1324$

Using the standard normal table:

$P(Z < -0.8529) \approx 0.1969, \quad P(Z < 0.1324) \approx 0.5527$

$P(5.29 < \bar{X} < 5.96) = P(-0.8529 < Z < 0.1324) = 0.5527 - 0.1969 = 0.3558$

Final Answers (Rounded to 4 Decimal Places)

• $P(5.29 < X < 5.96) = 0.1294$
• $P(5.29 < \bar{X} < 5.96) = 0.3558$

Do you want a more detailed explanation or any clarifications?

Related Questions:

1. What is the probability that a randomly selected patron spends more than $7.00 on lunch?
2. What is the probability that a group of 9 patrons has an average spending of more than $6.50?
3. How does increasing the sample size affect the probability calculations?
4. If a student has only $4.50, what is the probability that they can afford lunch?
5. What is the probability that the average spending for 25 patrons is less than $5.50?

Tip:

The standard deviation of the sample mean ($\bar{X}$) is called the standard error and is always smaller than the population standard deviation. This makes the sample mean more stable as the sample size increases.