Question 1186870
Here's the solution:

**1. a. Distribution of X:**

X represents the amount of syrup used by a single person.  We're given that it's normally distributed with a mean (μ) of 63 mL and a standard deviation (σ) of 11 mL.  So:

X ~ N(63, 11)

**1. b. Distribution of x̄:**

x̄ represents the average amount of syrup used by a sample of 50 people.  The distribution of the sample mean is also normal.  The mean of x̄ is the same as the population mean (μ = 63). The standard deviation of x̄ (also called the standard error) is the population standard deviation (σ = 11) divided by the square root of the sample size (n = 50):

Standard Error = σ / √n = 11 / √50 ≈ 1.5556

So, x̄ ~ N(63, 1.5556)

**1. c. Probability for a single individual:**

We want to find P(61.8 < X < 64).  First, convert these values to z-scores:

z₁ = (61.8 - 63) / 11 ≈ -0.1091
z₂ = (64 - 63) / 11 ≈ 0.0909

Now, look up these z-scores in a standard normal distribution table (or use a calculator) to find the corresponding probabilities:

P(z < -0.1091) ≈ 0.4565
P(z < 0.0909) ≈ 0.5362

P(61.8 < X < 64) = P(z < 0.0909) - P(z < -0.1091) = 0.5362 - 0.4565 ≈ 0.0797

**1. d. Probability for the average of 50 people:**

We want to find P(61.8 < x̄ < 64). Use the same z-score formula, but with the standard error of the mean:

z₁ = (61.8 - 63) / 1.5556 ≈ -0.7712
z₂ = (64 - 63) / 1.5556 ≈ 0.6422

P(z < -0.7712) ≈ 0.2200
P(z < 0.6422) ≈ 0.7396

P(61.8 < x̄ < 64) = P(z < 0.6422) - P(z < -0.7712) = 0.7396 - 0.2200 ≈ 0.5196

**Answers:**

a. X ~ N(63, 11)
b. x̄ ~ N(63, 1.5556)
c. 0.0797
d. 0.5196