To solve this problem, we need to calculate the probability that the mean of the 12 randomly selected scores is between 30 and 32, given that Mike's scores are approximately normally distributed with a mean of 31.6 and a standard deviation of 3.2.

Steps:

1. Given data:

   • Mean (μ) = 31.6
   • Standard deviation (σ) = 3.2
   • Sample size (n) = 12
   • We need to find the probability that the sample mean is between 30 and 32.

2. Standard deviation of the sample mean: The standard deviation of the sample mean (also called the standard error) is given by the formula:

   $\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{3.2}{\sqrt{12}}$

3. Calculate Z-scores: We now calculate the Z-scores for 30 and 32 using the formula:

   $Z = \frac{X - \mu}{\text{Standard Error}}$

   where $X$ is the value we are interested in (30 and 32).

4. Look up the Z-scores in a standard normal distribution table or use a calculator to find the corresponding probabilities.

Let's start with the calculations.

I'll calculate the standard error first. math

Given data

sigma = 3.2 # standard deviation n = 12 # sample size

Calculate standard error

standard_error = sigma / math.sqrt(n) standard_errorThe standard error of the sample mean is approximately 0.924.

Next, let's calculate the Z-scores for 30 and 32.

Using the formula:

$Z = \frac{X - \mu}{\text{Standard Error}}$

We will calculate the Z-scores for $X = 30$ and $X = 32$, where $\mu = 31.6$. Mean value mu = 31.6

Calculate Z-scores for 30 and 32

z_score_30 = (30 - mu) / standard_error z_score_32 = (32 - mu) / standard_error

z_score_30, z_score_32The Z-scores are:

• For $X = 30$, the Z-score is approximately $-1.73$.
• For $X = 32$, the Z-score is approximately $0.43$.

Now, we can use these Z-scores to find the probabilities corresponding to them from the standard normal distribution table (or using a calculator). Let's look up the probabilities for each Z-score. scipy.stats as stats

Find probabilities corresponding to the Z-scores

prob_30 = stats.norm.cdf(z_score_30) # Probability for Z = -1.73 prob_32 = stats.norm.cdf(z_score_32) # Probability for Z = 0.43

Probability that the sample mean is between 30 and 32

prob_between_30_and_32 = prob_32 - prob_30

prob_30, prob_32, prob_between_30_and_32The probabilities are:

• The probability for $Z = -1.73$ (corresponding to $X = 30$) is approximately 0.0416.
• The probability for $Z = 0.43$ (corresponding to $X = 32$) is approximately 0.6675.
• Therefore, the probability that the sample mean is between 30 and 32 is approximately $0.626$ or 62.6%.

If you need further details or clarification, feel free to ask!

Here are some related questions to explore further:

1. What happens to the probability if the sample size were increased?
2. How can you calculate confidence intervals using the same principles?
3. What is the relationship between Z-scores and standard deviation in normal distributions?
4. How do you find the Z-scores for non-standard normal distributions?
5. What impact does the standard deviation have on the probability of sample means?

Tip: Always ensure you understand the concept of standard error as it helps in determining how much variability there is in the sample mean compared to the population mean.