Snake Lengths: Probability Above 16.6 Inches?

Hey Plastik Magazine readers! Let's dive into a cool math problem today that involves our slithery friends: snakes! We're going to figure out the percentage of snakes that are longer than a certain length, given some statistical information. So, grab your thinking caps, and let's get started!

Understanding Normal Distribution in Snake Lengths

First off, we're told that the lengths of a particular type of snake follow a normal distribution. What does that even mean, right? Well, imagine a bell curve – that's basically what a normal distribution looks like. In our case, this bell curve represents the distribution of snake lengths. The peak of the curve shows the average length (the mean), and how spread out the curve is tells us about the standard deviation. Think of standard deviation as the average distance each snake's length is from the average length of all the snakes.

In this problem, the mean ($\\mu$) is 15 inches, and the standard deviation (\$\sigma\\$) is 0.8 inches. So, the average snake length is 15 inches, and most snake lengths will be within a certain range around this average. A smaller standard deviation would mean the snake lengths are clustered more closely around the mean, while a larger standard deviation means they're more spread out. Given the mean and standard deviation of the snake lengths, figuring out the probability or likelihood of certain lengths involves using the properties of normal distributions. We’re essentially trying to find out how much of the area under the bell curve lies to the right of 16.6 inches. This area corresponds to the percentage of snakes longer than 16.6 inches. The normal distribution is incredibly useful in statistics because it pops up in so many real-world situations, from heights and weights to test scores and, yes, even snake lengths! It allows us to make predictions and understand the variation within a population.

Calculating the Z-Score for Snake Lengths

Okay, so we know our snake lengths are normally distributed, and we want to find the percentage of snakes longer than 16.6 inches. How do we do that? This is where the z-score comes in handy. The z-score is a way of standardizing a value within a normal distribution. It tells us how many standard deviations a particular value is away from the mean. A positive z-score means the value is above the mean, and a negative z-score means it's below the mean.

To calculate the z-score, we use a simple formula:

$z = (x - \\mu) / \\sigma$

Where:

• x is the value we're interested in (in this case, 16.6 inches)
• \$\mu\\$ is the mean (15 inches)
• \$\sigma\\$ is the standard deviation (0.8 inches)

Let's plug in the numbers:

$z = (16.6 - 15) / 0.8 = 1.6 / 0.8 = 2$

So, the z-score for 16.6 inches is 2. This means that a snake length of 16.6 inches is 2 standard deviations above the average snake length. Why is this important? Well, the z-score allows us to use a standard normal distribution table (or a calculator) to find the probability associated with that z-score. In essence, by converting our specific problem (snake lengths) into a standard form (z-score), we can use readily available tools to find the answer. Remember, the z-score is a crucial step in working with normal distributions because it bridges the gap between specific values and the probabilities we’re trying to find. This is also extremely helpful when comparing different distributions with different means and standard deviations, as the Z-score acts as a common yardstick.

Finding the Probability Using the Z-Table for Snake Lengths

Now that we've calculated the z-score (which is 2), the next step is to find the corresponding probability. This is where the z-table (also known as the standard normal distribution table) comes in. A z-table is a table that shows the cumulative probability associated with a given z-score. In other words, it tells us the probability of a value being less than or equal to a certain z-score.

Most z-tables will give you the area to the left of the z-score. However, we're interested in the percentage of snakes longer than 16.6 inches, so we need to find the area to the right of our z-score (z = 2). To do this, we'll subtract the probability from the z-table from 1 (since the total area under the curve is 1, representing 100%).

Looking up a z-score of 2 in the z-table, we find a probability of approximately 0.9772. This means that about 97.72% of snakes are shorter than 16.6 inches.

Now, we subtract this from 1:

$1 - 0.9772 = 0.0228$

This gives us 0.0228, which, when converted to a percentage, is 2.28%. So, approximately 2.28% of the snakes are longer than 16.6 inches. When using the z-table, it’s crucial to correctly interpret the values and determine whether you need to find the area to the left or right of the z-score, or perhaps even between two z-scores. Understanding this relationship ensures accurate probability calculations.

Determining the Correct Answer for Snake Lengths

Okay, we've crunched the numbers, and we've found that approximately 2.28% of the snakes are longer than 16.6 inches. Now, let's look back at the answer choices:

A. 0.3% B. 2.5% C. 3.5% D. 5%

Our calculated percentage (2.28%) is closest to option B, which is 2.5%. So, the correct answer is B. 2.5%.

Therefore, approximately 2.5% of the snakes are longer than 16.6 inches. It's pretty cool how we can use statistics to understand things like the distribution of snake lengths, right? The key to solving problems like this is understanding the concepts of normal distribution, z-scores, and how to use the z-table. Once you've got those down, you can tackle all sorts of probability problems! And remember, math isn't just about formulas and numbers; it's a powerful tool for understanding the world around us, even the slithery parts!

So there you have it, folks! We've successfully navigated the world of snake lengths and normal distributions. Hope you enjoyed this mathematical adventure. Keep those brains buzzing, and we'll catch you in the next one!