Let's solve the problems step-by-step.

Problem 1: Probability of a Single Value

Given a normal distribution with a mean $\mu = 113.8$ and standard deviation $\sigma = 22.9$, we need to find the probability that a single randomly selected value $X$ falls between 109 and 113.8:

$P(109 < X < 113.8)$

Step 1: Calculate Z-scores

The Z-score formula is:

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

For $X = 109$:

$Z_1 = \frac{109 - 113.8}{22.9} = \frac{-4.8}{22.9}$

For $X = 113.8$:

$Z_2 = \frac{113.8 - 113.8}{22.9} = 0$

Let's calculate these values.

Problem 2: Probability of a Sample Mean

Now, find the probability that a sample of size $n = 129$ is randomly selected with a mean between 109 and 113.8:

$P(109 < M < 113.8)$

Step 1: Calculate the Standard Error (SE) The standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}} = \frac{22.9}{\sqrt{129}}$

Step 2: Calculate Z-scores for the sample mean

For $M = 109$:

$Z_3 = \frac{109 - 113.8}{SE}$

For $M = 113.8$:

$Z_4 = \frac{113.8 - 113.8}{SE} = 0$

I'll calculate these values now.Here are the calculations for each problem:

Problem 1: Probability of a Single Value

• Z-score for $X = 109$: $Z_1 = -0.2096$
• Z-score for $X = 113.8$: $Z_2 = 0$
• Probability $P(109 < X < 113.8) = 0.0830$

Problem 2: Probability of a Sample Mean

• Standard Error (SE): $SE = 2.0165$ (calculated internally)
• Z-score for $M = 109$: $Z_3 = -2.3807$
• Z-score for $M = 113.8$: $Z_4 = 0$
• Probability $P(109 < M < 113.8) = 0.4914$

Would you like further explanation on any part of the solution?

Here are some related questions:

1. What is the importance of standard error in probability calculations?
2. How do Z-scores relate to probabilities in a normal distribution?
3. Why is the probability of a sample mean different from that of a single value?
4. How would the probabilities change if the sample size increased?
5. What would the probabilities be if the range was adjusted slightly?

Tip: Always double-check the Z-score calculations when dealing with normal distributions to ensure accuracy!