Let's break down each part of the question step by step:

(a) The distribution of students' grades is may or may not be normal.

• The distribution of individual students' grades is described as unimodal and skewed to the left, which suggests that it is not normal. Since it's skewed to the left, it's unlikely to be normal. However, the shape of the individual distribution does not necessarily dictate the shape of the sampling distribution of the sample mean.

(b) If $n = 30$ students are selected at random, the distribution of the sample mean grade is approximately normal with a mean of 60 and a standard deviation of 3.0305.

• For $n = 30$, the Central Limit Theorem tells us that the distribution of the sample mean will be approximately normal, even if the original distribution of the individual grades is not normal.

• The mean of the sample mean is the same as the population mean $\mu = 60$.

• The standard deviation of the sample mean (also known as the standard error) is calculated as:

  $\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{16.6}{\sqrt{30}} \approx 3.0305$

(c) The probability that the sample mean grade for these 30 students is less than 63.0 is 0.8389.

• To calculate the probability, we need to standardize the value 63.0 using the z-score formula:

  $Z = \frac{X - \mu}{\text{Standard Error}} = \frac{63.0 - 60}{3.0305} \approx 0.989$

  Using a standard normal distribution table (or a calculator), the probability corresponding to $Z = 0.989$ is approximately 0.8389.

(d) If $n = 30$ students are selected at random, the distribution of the sample total grade is approximately normal with a mean of 1800 and a standard deviation of 90.9158.

• The mean of the total grade is the sum of the sample means, so:

  $\text{Mean of total} = n \times \mu = 30 \times 60 = 1800$

• The standard deviation of the total grade is calculated as:

  $\text{Standard deviation of total} = \text{Standard Error} \times \sqrt{n} = 3.0305 \times \sqrt{30} \approx 90.9158$

(e) The probability that the total grade for these 30 students is less than 1890.0 is 0.8389.

• To calculate the probability, we first standardize the total grade of 1890.0 using the z-score formula:

  $Z = \frac{X - \text{Mean of total}}{\text{Standard deviation of total}} = \frac{1890.0 - 1800}{90.9158} \approx 0.989$

  Using a standard normal distribution table, the probability corresponding to $Z = 0.989$ is approximately 0.8389.

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Would you like further details on any of the calculations, or is there another concept you'd like to explore?

Additional Questions:

1. How do I calculate the z-score for any given value in a normal distribution?
2. What is the Central Limit Theorem and how does it apply to sampling distributions?
3. How can I interpret a z-score in terms of probability?
4. Why does the standard deviation of the sample mean decrease as the sample size increases?
5. How can I calculate probabilities for other normal distributions or use a calculator to do it?

Tip: When dealing with sample statistics, always ensure you're using the sample size correctly in your calculations for the standard error and standard deviation of the sample total.