Normal Distribution Examples: Problems, Solutions & Graphs

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Normal Distribution Examples: Problems, Solutions & Graphs

Hey guys! Ever wondered how the Normal Distribution works in real life? It's not just a theoretical concept; it pops up everywhere! Let's break down two examples with step-by-step solutions and graphs, making it super easy to understand.

Example 1: Heights of Adults

Okay, so imagine we're looking at the heights of adult men. We know that heights often follow a normal distribution pattern. Let's say the average height (mean) is 5'10" (that’s about 70 inches or 178 cm), and the standard deviation is about 3 inches (7.62 cm). This means most men cluster around that average height, and the further you get from 5'10", the fewer men you'll find at those heights.

Problem Statement

What's the probability that a randomly selected man is taller than 6' (72 inches or 183 cm)?

Step-by-Step Solution

  1. Define the Random Variable: Let X be the height of a randomly selected man. We know that X follows a normal distribution with a mean (μ) of 70 inches and a standard deviation (σ) of 3 inches. We write this as X ~ N(70, 3^2).

  2. Standardize the Value: To find the probability, we need to convert the height of 72 inches into a z-score. The z-score tells us how many standard deviations away from the mean our value is. The formula for the z-score is:

    z = (X - μ) / σ

    In our case:

    z = (72 - 70) / 3 = 2 / 3 ≈ 0.67

    So, a height of 72 inches is 0.67 standard deviations above the mean.

  3. Find the Probability: Now, we need to find the probability that a z-score is greater than 0.67. We can use a z-table (also known as a standard normal table) or a calculator with statistical functions to find this probability. A z-table gives us the area to the left of the z-score, so we need to subtract the z-table value from 1 to find the area to the right.

    Looking up 0.67 in a z-table, we find a value of approximately 0.7486. This means that P(Z < 0.67) = 0.7486.

    Therefore, the probability that a randomly selected man is taller than 6' is:

    P(X > 72) = P(Z > 0.67) = 1 - P(Z < 0.67) = 1 - 0.7486 = 0.2514

    So, there's about a 25.14% chance that a randomly selected man is taller than 6 feet.

  4. Interpretation: This result tells us that approximately 25.14% of adult men are taller than 6 feet, given our assumed normal distribution of heights with a mean of 5'10" and a standard deviation of 3 inches.

Graph

Imagine a bell-shaped curve. The peak of the curve is at 70 inches (the mean). The area under the curve to the right of 72 inches represents the probability we just calculated (0.2514). The x-axis represents height in inches, and the y-axis represents the probability density. The z-score of 0.67 marks the point 72 inches relative to the mean in standard deviations. You'd shade the area to the right of z = 0.67 to visually represent the probability.

Example 2: Exam Scores

Let's say you've just taken a really important exam. The scores on this exam also happen to follow a normal distribution. The average score (mean) is 75, and the standard deviation is 7.

Problem Statement

What percentage of students scored between 80 and 90?

Step-by-Step Solution

  1. Define the Random Variable: Let X be the score of a randomly selected student. X follows a normal distribution with a mean (μ) of 75 and a standard deviation (σ) of 7. We write this as X ~ N(75, 7^2).

  2. Standardize the Values: We need to convert both 80 and 90 into z-scores:

    For 80:

    z1 = (80 - 75) / 7 = 5 / 7 ≈ 0.71

    For 90:

    z2 = (90 - 75) / 7 = 15 / 7 ≈ 2.14

    So, a score of 80 is approximately 0.71 standard deviations above the mean, and a score of 90 is approximately 2.14 standard deviations above the mean.

  3. Find the Probabilities: We need to find the probability that a z-score is between 0.71 and 2.14. We can use a z-table or a calculator.

    Looking up 0.71 in a z-table, we find a value of approximately 0.7611. This means P(Z < 0.71) = 0.7611.

    Looking up 2.14 in a z-table, we find a value of approximately 0.9838. This means P(Z < 2.14) = 0.9838.

    To find the probability that a student scored between 80 and 90, we subtract the probability of scoring less than 80 from the probability of scoring less than 90:

    P(80 < X < 90) = P(0.71 < Z < 2.14) = P(Z < 2.14) - P(Z < 0.71) = 0.9838 - 0.7611 = 0.2227

    So, about 22.27% of students scored between 80 and 90.

  4. Interpretation: This indicates that given the exam scores follow a normal distribution, roughly 22.27% of the students achieved scores within the range of 80 to 90.

Graph

Again, picture a bell curve. The peak is at 75 (the mean score). You'd shade the area under the curve between 80 and 90. The x-axis represents the exam scores, and the y-axis represents the probability density. The z-scores 0.71 and 2.14 mark the positions of 80 and 90, respectively, relative to the mean in terms of standard deviations.

Key Takeaways

  • The normal distribution is everywhere: From heights to exam scores, it's a common pattern.
  • Z-scores are your friends: They help you standardize values and compare them across different normal distributions.
  • Z-tables (or calculators) are essential: They allow you to find probabilities associated with z-scores.
  • Understanding the problem is crucial: Always clearly define what you're trying to find.

So there you have it! Two examples of how the normal distribution works in the real world, along with step-by-step solutions and how to visualize them with graphs. Hope this helps you guys get a better handle on this important statistical concept! Remember to practice and you'll become a Normal Distribution pro in no time!