Module 4 Practice Problems
20 Comprehensive Problems • Covers All Module 4 Topics
How to Use These Practice Problems
- Work through problems on paper before checking solutions
- Use the hints if you're stuck
- Problems are organized by topic and difficulty
- You'll need a z-table for many problems (provided in lessons)
- Show all your work - understanding the process is key!
Problems 1-5: Normal Distribution Basics & Empirical Rule
Problem 1 Identifying Normal Distribution Easy
List three key characteristics that define a normal distribution.
Solution:
Three key characteristics:
- Bell-shaped and symmetric around the mean
- Mean = Median = Mode (all at the center)
- Follows Empirical Rule: 68% within 1σ, 95% within 2σ, 99.7% within 3σ
Additional properties: Continuous curve, tails approach but never touch the x-axis, total area under curve = 1
Problem 2 Empirical Rule Application Easy
Test scores are normally distributed with mean 75 and standard deviation 8. According to the Empirical Rule, approximately what percent of scores fall between 67 and 83?
Solution:
- Lower bound: 75 − 8 = 67 (1 SD below mean)
- Upper bound: 75 + 8 = 83 (1 SD above mean)
- By Empirical Rule: 68% of data falls within 1 SD
Answer: Approximately 68% of scores fall between 67 and 83
Problem 3 Two Standard Deviations Easy
Heights of adult women are normally distributed with μ = 64 inches and σ = 2.5 inches. What range contains approximately 95% of all heights?
Solution:
95% falls within 2 standard deviations:
Lower bound: μ − 2σ = 64 − 2(2.5) = 64 − 5 = 59 inches
Upper bound: μ + 2σ = 64 + 2(2.5) = 64 + 5 = 69 inches
Answer: 95% of heights fall between 59 and 69 inches
Problem 4 Three Standard Deviations Medium
IQ scores follow a normal distribution with mean 100 and standard deviation 15. What percent of people have IQ scores between 55 and 145?
Solution:
Check the boundaries:
Lower: 100 − 3(15) = 100 − 45 = 55
Upper: 100 + 3(15) = 100 + 45 = 145
These are exactly 3 standard deviations from the mean!
By Empirical Rule: 99.7% falls within 3σ
Answer: 99.7% of people have IQ scores between 55 and 145
Problem 5 Empirical Rule Percentages Medium
For a normal distribution with μ = 50 and σ = 5, what percentage of data falls ABOVE 60?
Solution:
60 = 50 + 2(5), so 60 is 2 SDs above the mean
By Empirical Rule: 95% falls within 2σ (between 40 and 60)
This means 5% falls outside this range
By symmetry: 2.5% below 40 and 2.5% above 60
Answer: 2.5% of data falls above 60
Problems 6-10: Calculating and Interpreting Z-scores
Problem 6 Basic Z-score Calculation Easy
Calculate the z-score for x = 85 when μ = 75 and σ = 10.
Solution:
z = (x − μ) / σ
z = (85 − 75) / 10
z = 10 / 10
z = 1.0
Answer: z = 1.0
Interpretation: 85 is exactly 1 standard deviation above the mean
Problem 7 Negative Z-score Easy
A student scores 68 on a test where μ = 80 and σ = 6. Calculate and interpret the z-score.
Solution:
z = (x − μ) / σ
z = (68 − 80) / 6
z = −12 / 6
z = −2.0
Answer: z = −2.0
Interpretation: The score of 68 is 2 standard deviations BELOW the mean - this is an unusually low score
Problem 8 Finding X from Z-score Medium
For a normal distribution with μ = 200 and σ = 25, find the value x that has a z-score of 1.5.
Solution:
Use: x = μ + z·σ
x = 200 + 1.5(25)
x = 200 + 37.5
x = 237.5
Answer: x = 237.5
Check: z = (237.5 − 200) / 25 = 37.5 / 25 = 1.5
Problem 9 Comparing Scores Medium
Sarah scored 88 on Test A (μ = 80, σ = 4) and 76 on Test B (μ = 70, σ = 3). On which test did she perform better relative to her classmates?
Solution:
Test A:
z = (88 − 80) / 4 = 8 / 4 = 2.0
Test B:
z = (76 − 70) / 3 = 6 / 3 = 2.0
Both z-scores are 2.0!
Answer: Sarah performed equally well (relative to classmates) on both tests
On both tests, she scored exactly 2 standard deviations above the mean
Problem 10 Unusual Values Hard
Birth weights are normally distributed with μ = 7.5 pounds and σ = 1.2 pounds. A baby weighing 4.9 pounds is born. Is this weight unusual? (Use the criterion that |z| > 2 indicates an unusual value)
Solution:
z = (x − μ) / σ
z = (4.9 − 7.5) / 1.2
z = −2.6 / 1.2
z ≈ −2.17
|z| = 2.17 > 2
Answer: YES, this weight is unusual
The baby weighs about 2.17 standard deviations below the mean - this would be considered a low birth weight requiring medical attention
Problems 11-15: Finding Probabilities
Problem 11 Left-tail Probability Medium
For a standard normal distribution, find P(Z < 1.25).
Solution:
Look up z = 1.25 in the z-table:
Row 1.2, Column .05 → 0.8944
Answer: P(Z < 1.25) = 0.8944 or 89.44%
About 89% of values fall below z = 1.25
Problem 12 Right-tail Probability Medium
For a standard normal distribution, find P(Z > 0.75).
Solution:
Step 1: Look up z = 0.75 → 0.7734
This is P(Z < 0.75)
Step 2: P(Z > 0.75) = 1 − P(Z < 0.75)
= 1 − 0.7734
= 0.2266
Answer: P(Z > 0.75) = 0.2266 or 22.66%
Problem 13 Between Two Values Medium
For a standard normal distribution, find P(−1.0 < Z < 1.5).
Solution:
P(−1.0 < Z < 1.5) = P(Z < 1.5) − P(Z < −1.0)
Step 1: P(Z < 1.5) = 0.9332 (from z-table)
Step 2: P(Z < −1.0) = 0.1587 (from z-table)
Step 3: 0.9332 − 0.1587 = 0.7745
Answer: P(−1.0 < Z < 1.5) = 0.7745 or 77.45%
Problem 14 Non-standard Normal Hard
SAT scores are normally distributed with μ = 1050 and σ = 200. What is the probability a randomly selected student scores below 900?
Solution:
Step 1: Convert to z-score
z = (900 − 1050) / 200 = −150 / 200 = −0.75
Step 2: Find P(Z < −0.75)
From z-table: P(Z < −0.75) = 0.2266
Answer: P(X < 900) = 0.2266 or 22.66%
About 23% of students score below 900
Problem 15 Between Two Non-standard Values Hard
Adult male heights are normally distributed with μ = 70 inches and σ = 3 inches. What percent of men are between 67 and 76 inches tall?
Solution:
Step 1: Convert to z-scores
z₁ = (67 − 70) / 3 = −3 / 3 = −1.0
z₂ = (76 − 70) / 3 = 6 / 3 = 2.0
Step 2: Find P(−1.0 < Z < 2.0)
P(Z < 2.0) = 0.9772
P(Z < −1.0) = 0.1587
Step 3: 0.9772 − 0.1587 = 0.8185
Answer: 81.85% of men are between 67 and 76 inches tall
Problems 16-20: Finding Percentiles and Applications
Problem 16 Finding a Z-score from Percentile Medium
Find the z-score that corresponds to the 75th percentile.
Solution:
75th percentile means P(Z < z) = 0.75
Look inside z-table for 0.7500:
Closest value is 0.7486 at z = 0.67
Or 0.7517 at z = 0.68
Answer: z ≈ 0.67 or 0.68
Most commonly: z = 0.674 for exactly 75th percentile
Problem 17 Finding a Value from Percentile Medium
Test scores are normally distributed with μ = 500 and σ = 100. What score represents the 90th percentile?
Solution:
Step 1: Find z for P(Z < z) = 0.90
From z-table: z ≈ 1.28
Step 2: Convert to x-value
x = μ + z·σ
x = 500 + 1.28(100)
x = 500 + 128
x = 628
Answer: The 90th percentile is a score of 628
90% of students score below 628
Problem 18 Top Percent Hard
IQ scores are normally distributed with μ = 100 and σ = 15. What IQ score puts someone in the top 10% of the population?
Solution:
Top 10% = 90th percentile
Step 1: Find z where P(Z < z) = 0.90
z ≈ 1.28
Step 2: Convert to IQ score
x = 100 + 1.28(15)
x = 100 + 19.2
x = 119.2
Answer: An IQ of approximately 119 puts someone in the top 10%
Problem 19 Quality Control Application Hard
A machine fills bottles with a mean of 16.0 oz and standard deviation of 0.2 oz (normally distributed). Bottles with less than 15.6 oz or more than 16.4 oz are rejected. What percentage of bottles are rejected?
Solution:
Step 1: Convert to z-scores
z₁ = (15.6 − 16.0) / 0.2 = −0.4 / 0.2 = −2.0
z₂ = (16.4 − 16.0) / 0.2 = 0.4 / 0.2 = 2.0
Step 2: Find rejection probabilities
P(Z < −2.0) = 0.0228
P(Z > 2.0) = 1 − 0.9772 = 0.0228
Step 3: Total rejected
0.0228 + 0.0228 = 0.0456
Answer: 4.56% of bottles are rejected
Note: This is the 5% outside ±2σ from Empirical Rule!
Problem 20 Medical Screening Application Hard
Blood pressure readings are normally distributed with μ = 120 and σ = 15. A reading above 150 is considered high and requires treatment. If a clinic sees 500 patients, approximately how many will need treatment?
Solution:
Step 1: Convert to z-score
z = (150 − 120) / 15 = 30 / 15 = 2.0
Step 2: Find P(Z > 2.0)
P(Z > 2.0) = 1 − P(Z < 2.0)
= 1 − 0.9772
= 0.0228
Step 3: Calculate number of patients
500 × 0.0228 = 11.4 patients
Answer: Approximately 11-12 patients will need treatment
This represents about 2.3% of patients
Great Work!
You've completed all 20 practice problems! Make sure you understand each solution before moving on to the quiz.