The problem in the image involves finding probabilities using the normal distribution. Let's break it down:

Given:

• The population has a normal distribution with:
  • Mean ($\mu$) = 35.6
  • Standard deviation ($\sigma$) = 51.7
  • Sample size ($n$) = 16

Part (a):

Find the probability that a single randomly selected value $X$ is less than 12.3: $P(X < 12.3)$ For a single value, use the formula for the standard normal distribution: $Z = \frac{X - \mu}{\sigma}$ Substitute the given values: $Z = \frac{12.3 - 35.6}{51.7}$

Part (b):

Find the probability that the sample mean $M$ for a sample of size $n = 16$ is less than 12.3: $P(M < 12.3)$ For the sample mean, use the standard error of the mean: $\sigma_{\text{mean}} = \frac{\sigma}{\sqrt{n}} = \frac{51.7}{\sqrt{16}} = \frac{51.7}{4} = 12.925$ Then calculate the $Z$-score for the sample mean: $Z = \frac{12.3 - 35.6}{12.925}$

I will now calculate the $Z$-scores and use them to find the probabilities from the standard normal distribution table.### Results:

1. Part (a): The $Z$-score for $P(X < 12.3)$ is approximately $-0.451$, and the corresponding probability is: $P(X < 12.3) \approx 0.3261$

2. Part (b): The $Z$-score for $P(M < 12.3)$ is approximately $-1.803$, and the corresponding probability is: $P(M < 12.3) \approx 0.0357$

Let me know if you would like further details or clarifications!

Here are some related questions you can explore:

1. How do you interpret the Z-score in a normal distribution?
2. What is the difference between population standard deviation and standard error of the mean?
3. How would the probability change if the sample size increased?
4. What role does the central limit theorem play in this problem?
5. How do you find probabilities for non-standard normal distributions?

Tip: When dealing with sample means, always use the standard error of the mean to account for the reduced variability due to larger sample sizes.