20.6 - Example Problems
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Understanding Example 1
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Today we're solving a problem involving the Normal Distribution. Let's start with Example 1, where we have students' marks normally distributed with a mean of 70 and a standard deviation of 10. Can anyone remind us why we need the Z-score?
The Z-score helps standardize the values so we can find probabilities!
Exactly! Now, let's convert our value of 85 into a Z-score. Who can help me with that?
We use the formula Z equals (X minus mu) divided by sigma, so Z equals (85 minus 70) divided by 10, which is 1.5!
Great job! Now, consulting the Z-table, what probability do we find for Z less than 1.5?
The Z-table tells us that P(Z < 1.5) is 0.9332.
Correct! So what's the final probability of a student scoring less than 85?
The probability is 0.9332 or 93.32%!
Excellent! This shows how to tackle a problem using Z-scores. Recapping: we identified our mean and standard deviation, converted to a Z-score, and used the Z-table to find probabilities.
Understanding Example 2
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Let's move on to Example 2, where we need to find the probability that a value lies between 60 and 80. What do we begin with?
We’ll first identify the mean and standard deviation!
Correct! We have μ = 70 and σ = 10. Now, who can help with converting 60 and 80 to Z-scores?
For 60, it's Z = (60 - 70) / 10 = -1, and for 80, Z = (80 - 70) / 10 = 1.
Excellent! What do we do next with these Z-scores?
We look up the Z-table: P(Z < 1) is 0.8413 and P(Z < -1) is 0.1587.
Good! Now how do we find the probability that a value is between 60 and 80?
We subtract the two probabilities: 0.8413 - 0.1587 = 0.6826.
Exactly right! The final answer is 68.26%. To summarize today’s discussion, we reviewed the steps of converting observed values into Z-scores and used the Z-table to find the relevant probabilities.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
In this section, we cover two example problems related to the Normal Distribution, detailing the processes for calculating probabilities. These examples illustrate conversions to Z-scores and the use of Z-tables.
Detailed
Example Problems
This section explores two practical examples to solidify the understanding of the Normal Distribution and associated calculations.
Understanding the Problems
Both examples focus on the fundamental steps involved in solving problems related to Normal Distribution: identifying the mean and standard deviation, calculating Z-scores, and referring to Z-tables to find probabilities.
Example 1:
Problem: The marks of students in a class are normally distributed with a mean of 70 and a standard deviation of 10. What is the probability that a student scores less than 85?
Solution:
1. Identifying Variables: Mean (μ) = 70, Standard Deviation (σ) = 10, X = 85.
2. Convert to Z-score:
\[ Z = \frac{X - μ}{σ} = \frac{85 - 70}{10} = 1.5 \]
3. Using the Z-table:
\[ P(Z < 1.5) = 0.9332 \]
Final Answer: Probability = 0.9332 or 93.32%.
Example 2:
Problem: Find the probability that a value lies between 60 and 80.
Solution:
1. Identifying Variables: Mean (μ) = 70, Standard Deviation (σ) = 10.
2. Convert both values to Z-scores:
- For 60: \[ Z = \frac{60 - 70}{10} = -1 \]
- For 80: \[ Z = \frac{80 - 70}{10} = 1 \]
3. Using the Z-table:
- \[ P(Z < 1) = 0.8413 \]
- \[ P(Z < -1) = 0.1587 \]
- \[ P(60 < X < 80) = 0.8413 - 0.1587 = 0.6826 \]
Final Answer: 68.26%
These examples illustrate how to apply the steps systematically to solve problems relating to the Normal Distribution, highlighting the importance of Z-scores and tables.
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Example 1: Probability of Scoring Less than 85
Chapter 1 of 2
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Chapter Content
Example 1:
The marks of students in a class are normally distributed with a mean of 70 and standard deviation of 10. What is the probability that a student scores less than 85?
Solution:
• Given: 𝜇 = 70, 𝜎 = 10, 𝑋 = 85
• Convert to Z:
85− 70
𝑍 = = 1.5
10
• From Z-table: 𝑃(𝑍 < 1.5) = 0.9332
Answer: Probability = 0.9332 or 93.32%
Detailed Explanation
In this example, we begin by identifying the known values: the mean (𝜇) is 70, the standard deviation (𝜎) is 10, and we want to find out the probability that a student scores less than 85. To convert the raw score (85) to a Z-score, we subtract the mean from the score and divide by the standard deviation:
Z = (X - μ) / σ = (85 - 70) / 10 = 1.5.
Next, we use the Z-table to find the probability corresponding to a Z-score of 1.5, which gives us a probability of 0.9332. This means that there's a 93.32% chance that a student scores less than 85.
Examples & Analogies
Imagine you're in a classroom where the students' test scores follow a bell curve. If the average score is 70, and one student scores 85, we can find out how many students likely scored less than this. By calculating a Z-score, we can understand how exceptional that score is within the context of the class. If 93 out of 100 students are expected to score lower than this student, it underscores the student's strong performance.
Example 2: Probability of Scoring Between 60 and 80
Chapter 2 of 2
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Chapter Content
Example 2:
Find the probability that a value lies between 60 and 80.
Solution:
• 𝜇 = 70, 𝜎 = 10
• Convert both values to Z:
60−70
o For 60: 𝑍 = = −1
1 10
80−70
o For 80: 𝑍 = = 1
2 10
From Z-table:
• 𝑃(𝑍 < 1) = 0.8413
• 𝑃(𝑍 < −1) = 0.1587
𝑃(60 < 𝑋 < 80)= 0.8413−0.1587 = 0.6826
Answer: 68.26%
Detailed Explanation
In this example, we are tasked with finding the probability that a student's score lies between 60 and 80. We start by noting the mean (𝜇) is still 70 and the standard deviation (𝜎) is 10. We convert both scores to Z-scores. For 60:
Z = (60-70) / 10 = -1, and for 80:
Z = (80-70) / 10 = 1.
Now, we reference the Z-table to determine the probabilities for these Z-scores: P(Z < 1) gives us 0.8413 and P(Z < -1) gives us 0.1587. To find the probability between these two scores, we subtract the smaller probability from the larger one: 0.8413 - 0.1587 = 0.6826, which indicates a 68.26% probability that a student's score will fall between 60 and 80.
Examples & Analogies
Think of a game where scoring between certain points matters. If the average score is set at 70, we want to know the chances of a player scoring between 60 and 80 points. Calculating Z-scores provides insights into how likely it is for players to achieve scores within that range, giving teams a better strategy to adapt their plays. If about 68 out of 100 players can be expected to score within this range, it suggests a competitive atmosphere but with room for strong performances.
Key Concepts
-
Normal Distribution: Describes data that clusters around a central value with symmetrical properties.
-
Z-score: A tool for standardization allowing comparison across different data sets.
-
Z-table: Provides cumulative probabilities for standard normal distribution based on Z-scores.
Examples & Applications
Example 1: Probability of a student scoring less than 85 is 93.32%.
Example 2: Probability of a value lying between 60 and 80 is 68.26%.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
In a bell curve, high and low, The mean is the center, that's how we know.
Stories
Imagine a teacher collecting test scores that shape a perfect bell. Now, Z-scores help us find where each score fits in relation to the average score, just like students sitting in a classroom based on their exam performance!
Memory Tools
Z for Zero means Standard; Calculate Z, find where you stand.
Acronyms
P.A.Z.
Probability
Area
Z-score.
Flash Cards
Glossary
- Normal Distribution
A continuous probability distribution that is symmetric about the mean, characterized by its bell-shaped curve.
- Zscore
A statistical measurement that describes a value's relation to the mean of a group of values, calculated as the number of standard deviations away from the mean.
- Ztable
A table that shows the cumulative probabilities of a standard normal distribution for given Z-scores.
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