91Ó°ÊÓ

The concept of standard deviation is central to understanding how data varies around the mean. In this exercise, the standard deviation \(\sigma\) is provided as 1.20 for the number of courses a student registers for. Standard deviation measures the average distance of each data point from the mean of the dataset. A smaller standard deviation indicates that the data points are close to the mean, while a larger standard deviation indicates that the data is more spread out.

To determine which \(x\) values fall more than one standard deviation below the mean, subtract the standard deviation from the mean: \(\mu - \sigma = 4.66 - 1.20 = 3.46\). Thus, values like 1, 2, and 3 fall more than 1 standard deviation below the mean. Similarly, to find values that are more than two standard deviations away, calculate \(\mu - 2\sigma = 2.26\) and \(\mu + 2\sigma = 7.06\). This identifies the values 1 and 2 as more than two standard deviations below the mean, and 7 as more than two above.

Calculating the mean of a probability distribution helps us to understand the central or expected value of the data. In probability distributions like the one given, the mean \(\mu\) is calculated by multiplying each possible value \(x\) by its probability \(p(x)\) and summing these products:

• \(x = 1\), \(p(x) = 0.02\)
• \(x = 2\), \(p(x) = 0.03\)
• ... and so forth for each value

We find \(\mu\) as follows:\[ \mu = (1 \times 0.02) + (2 \times 0.03) + (3 \times 0.09) + (4 \times 0.25) + (5 \times 0.40) + (6 \times 0.16) + (7 \times 0.05) = 4.66 \]Thus, the average number of courses students register for is 4.66. This value acts as a central reference point to compare other data points within the distribution.

Probability calculations help us understand the likelihood of different events within a given distribution. In this exercise, we need to consider probabilities for specific conditions based on standard deviations.

• For part (a), the probability of \(x\) being more than 1 standard deviation below the mean involves summing the probabilities for values 1, 2, and 3: \(0.02 + 0.03 + 0.09 = 0.14\).
• For part (b), to find the probability of \(x\) being more than 2 standard deviations away from the mean, sum the probabilities for \(x = 1\), \(x = 2\), and \(x = 7\): \(0.02 + 0.03 + 0.05 = 0.10\).

These calculations tell us how likely it is for a student to fall into extreme categories in terms of the number of courses, either significantly less or more than the average. Understanding these probabilities helps in assessing risk and making informed decisions.