Understanding Normal Distribution and Test Scores

The problem describes a scenario involving test scores of 360 students that follow a normal distribution. We are given the mean score and the standard deviation. Our goal is to determine the percentage of students whose scores fall outside a specific range of scores.

Key Information Provided:

• Total number of students (N): 360
• Mean of scores ($\mu$): 45 points
• Standard deviation of scores ($\sigma$): 9 points
• Range of scores to consider: 36 - 54 points

Analyzing the Given Score Range

The score range provided is from 36 to 54. Let's see how this range relates to the mean and standard deviation of the normal distribution.

• Lower limit of the range: 36
• Upper limit of the range: 54

We can express these limits in terms of the mean ($\mu$) and standard deviation ($\sigma$):

• Lower limit: $36 = 45 - 9$. This is exactly $\mu - \sigma$.
• Upper limit: $54 = 45 + 9$. This is exactly $\mu + \sigma$.

So, the range of scores from 36 to 54 corresponds to the interval $(\mu - \sigma, \mu + \sigma)$ in this normal distribution.

Applying the Empirical Rule for Normal Distributions

A fundamental property of a normal distribution, often referred to as the Empirical Rule or the 68-95-99.7 Rule, states the approximate percentage of data that falls within a certain number of standard deviations from the mean:

• Approximately 68% of the data falls within one standard deviation ($\pm 1\sigma$) of the mean.
• Approximately 95% of the data falls within two standard deviations ($\pm 2\sigma$) of the mean.
• Approximately 99.7% of the data falls within three standard deviations ($\pm 3\sigma$) of the mean.

Calculating Percentage of Students Within the Range

Since the range 36 - 54 corresponds to $(\mu - \sigma, \mu + \sigma)$, according to the Empirical Rule, approximately 68% of the students' scores should fall within this range.

Calculating Percentage of Students Outside the Range

The question asks for the percentage of students who scored outside the limits of scores 36 - 54. If 68% of students scored within this range, the remaining percentage must have scored outside this range.

Therefore, approximately 32% of the students scored outside the limits of 36 - 54 points.

Step-by-Step Solution Summary

1. Identify the mean ($\mu = 45$) and standard deviation ($\sigma = 9$) of the normal distribution.
2. Determine the relationship between the given score limits (36 and 54) and the mean and standard deviation.
   • $36 = 45 - 9 = \mu - \sigma$
   • $54 = 45 + 9 = \mu + \sigma$
   The range is $(\mu - \sigma, \mu + \sigma)$.
3. Recall the Empirical Rule for normal distributions, specifically that approximately 68% of data lies within one standard deviation of the mean.
4. Conclude that approximately 68% of students scored within the 36-54 range.
5. Calculate the percentage of students scoring outside the range by subtracting the percentage within the range from 100%. $$ 100\% - 68\% = 32\% $$

Thus, 32% of students scored outside the specified limits.

Revision Table: Normal Distribution Concepts

Additional Information: Beyond the Empirical Rule

While the Empirical Rule provides good approximations, especially for values exactly at $\pm 1\sigma$, $\pm 2\sigma$, and $\pm 3\sigma$, the exact percentages for any range in a normal distribution can be found using Z-scores and the standard normal distribution table (or statistical software). The Z-score measures how many standard deviations a data point is from the mean.

For this problem:

• For score $X = 36$: $Z = \frac{36 - 45}{9} = \frac{-9}{9} = -1$
• For score $X = 54$: $Z = \frac{54 - 45}{9} = \frac{9}{9} = 1$

So, the range 36-54 corresponds to Z-scores between -1 and +1. The standard normal distribution table shows that the area under the curve between Z = -1 and Z = +1 is approximately 0.6827, or 68.27%. This is very close to the 68% given by the Empirical Rule. The percentage outside this range would be $100\% - 68.27\% = 31.73\%$. For the purpose of this question and the given options, the 68% approximation is sufficient and leads directly to the answer 32%.