Normal Distribution: Finding Sample Mean Thresholds

Statistics & Probability 11th-12th Grade
Problem
The reading speed of second grade students in a large city is approximately normal with a mean of 88 words per minute, and a standard deviation of 10 words per minute. There is a 5% chance that the mean reading speed of a random sample of 25 second grade students will exceed what value?

What You Will Learn

  • How to apply the Central Limit Theorem to find the distribution of sample means
  • The distinction between population standard deviation and standard error
  • How to work backwards from probability to find critical values using z-scores
  • When sample means follow normal distributions regardless of population normality
  • Real-world interpretation of statistical thresholds and percentiles
The reading speed of second grade students in a large city is approximately normal with a mean of 88 words per...

Solution: Method 1 — The Standard Error Approach

This problem asks us to find a threshold value where there's only a 5% chance the sample mean exceeds it. We're working with the sampling distribution of the mean, not individual student scores.

Step 1 — Identify the sampling distribution parameters

For a sample of size n = 25 from a normal population:

Mean of sample means: μ_X̄ = μ = 88 wpm
Standard error: SE = σ/√n = 10/√25 = 10/5 = 2 wpm

The key insight: individual students vary with standard deviation 10 wpm, but sample means of 25 students only vary with standard error 2 wpm. Sample means cluster much closer to the population mean.

Step 2 — Find the critical z-value

We want P(X̄ > k) = 0.05, which means P(X̄ ≤ k) = 0.95. We need the 95th percentile of the standard normal distribution:

z₀.₀₅ = 1.645

Step 3 — Apply the z-score formula

Convert from the standard normal back to our sampling distribution:

z = (k - μ)/SE
1.645 = (k - 88)/2
k - 88 = 1.645 × 2
k - 88 = 3.29
k = 91.29 wpm
The sample mean will exceed 91.29 words per minute with 5% probability.

Solution: Method 2 — The Inverse Normal Approach

Instead of working through z-scores step by step, we can think of this as a single inverse normal calculation on the sampling distribution X̄ ~ N(88, 2²).

Step 1 — Set up the inverse normal problem

We want the value k such that only 5% of sample means exceed it. This is the 95th percentile of the sampling distribution:

k = InvNorm(0.95, μ = 88, σ = 2)

Step 2 — Calculate directly

Using the inverse normal formula:

k = μ + z₀.₀₅ × SE
k = 88 + 1.645 × 2
k = 88 + 3.29
k = 91.29 wpm

This method treats the sampling distribution as our "new normal" and finds its 95th percentile directly.

91.29 words per minute

Verification

Let's confirm that P(X̄ > 91.29) = 0.05 by computing the z-score:

z = (91.29 - 88)/2 = 3.29/2 = 1.645

Looking up z = 1.645 in the standard normal table: P(Z > 1.645) = 1 - 0.95 = 0.05 ✓

We can also verify our standard error calculation. With σ = 10 and n = 25:

SE = 10/√25 = 10/5 = 2 ✓

Common Pitfalls

✗ Using the population standard deviation instead of standard error
Wrong: k = 88 + 1.645 × 10 = 104.45 wpm

This treats individual scores and sample means as having the same variability. Sample means are much less variable than individual scores because extreme values average out.
✗ Confusing "5% exceed" with "95% exceed"
Wrong: Using z = -1.645 (5th percentile instead of 95th)
k = 88 + (-1.645) × 2 = 84.71 wpm

"5% chance of exceeding k" means k is at the 95th percentile, not the 5th percentile.
✗ Forgetting to apply the Central Limit Theorem
Wrong: Thinking we need n ≥ 30 for normality

Since the population is already normal, sample means follow a normal distribution for any sample size. The CLT's n ≥ 30 rule only applies when the population isn't normal.

The Underlying Pattern

This problem follows the general structure for finding critical values in sampling distributions:

Critical Value Formula:
k = μ ± z_α × (σ/√n)

Where:
• μ = population mean
• z_α = critical z-value for probability α
• σ/√n = standard error of the mean
• Use + for "upper tail" problems (exceed)
• Use - for "lower tail" problems (fall below)

This pattern appears in confidence intervals, hypothesis testing, and quality control. The key insight is that sample means have reduced variability compared to individual observations—they cluster around μ with spread determined by the standard error, not the population standard deviation.

Important: This formula only works when sample means are normally distributed (either population is normal, or n ≥ 30 by CLT).

Why This Shows Up in the Real World

  • Quality Control: Manufacturing processes monitor average measurements from batches to detect when production goes off-spec
  • Medical Testing: Clinical trials evaluate whether average treatment effects exceed meaningful thresholds
  • Educational Assessment: School districts track whether average test scores meet proficiency standards

What If?

1
Smaller Sample Size
The reading speed of second grade students is normal with mean 88 wpm and standard deviation 10 wpm. If we take a random sample of only 9 students, what value will the sample mean exceed with probability 5%?
Step 1 — Calculate the standard error

With n = 9: SE = σ/√n = 10/√9 = 10/3 = 3.33 wpm

Step 2 — Apply the critical value formula

k = μ + z₀.₀₅ × SE = 88 + 1.645 × 3.33 = 88 + 5.48 = 93.48 wpm

Step 3 — Verify

z = (93.48 - 88)/3.33 = 1.645

Answer: 93.48 wpm

Note: Smaller samples have larger standard errors, so the threshold is higher.

2
Find the Probability
Reading speeds are normal with mean 88 wpm and standard deviation 10 wpm. What is the probability that a random sample of 25 students has a mean reading speed between 86 and 90 wpm?
Step 1 — Set up the sampling distribution

X̄ ~ N(88, 2²) where SE = 10/√25 = 2

Step 2 — Calculate z-scores for both bounds

z₁ = (86 - 88)/2 = -1
z₂ = (90 - 88)/2 = 1

Step 3 — Find the probability between z-scores

P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826

Answer: 68.26% or about 68%

This is the familiar "68% within one standard error" rule.

3
Different Confidence Level
Reading speeds are normal with mean 88 wpm and standard deviation 10 wpm. There is a 1% chance that the mean reading speed of 25 students will exceed what value?
Step 1 — Find the 99th percentile z-score

For 1% in upper tail: z₀.₀₁ = 2.326

Step 2 — Calculate the standard error

SE = 10/√25 = 2 wpm

Step 3 — Apply the formula

k = 88 + 2.326 × 2 = 88 + 4.65 = 92.65 wpm

Step 4 — Verify

z = (92.65 - 88)/2 = 2.326

Answer: 92.65 wpm

4
Reverse Engineering
Reading speeds are normal with mean 88 wpm. From a sample of 25 students, we know there's a 5% chance the sample mean exceeds 92 wpm. What is the population standard deviation?
Step 1 — Set up the equation

1.645 = (92 - 88)/(σ/√25)

Step 2 — Solve for σ

1.645 = 4/(σ/5)
1.645 = 20/σ
σ = 20/1.645 = 12.16

Step 3 — Verify

SE = 12.16/√25 = 2.43
k = 88 + 1.645 × 2.43 = 92

Answer: σ = 12.16 wpm

Frequently Asked Questions

How do you find critical values for sample means using the central limit theorem? +
Calculate the standard error (SE = σ/√n), find the z-score for your probability (like z = 1.645 for 95th percentile), then solve k = μ + z × SE. In this problem, SE = 10/√25 = 2, so k = 88 + 1.645 × 2 = 91.29 wpm.
What's the difference between population standard deviation and standard error? +
Population standard deviation (σ) measures spread of individual values. Standard error (SE = σ/√n) measures spread of sample means. Here, individual students vary by σ = 10 wpm, but sample means of 25 students only vary by SE = 2 wpm.
When can you use normal distribution for sample means? +
By the Central Limit Theorem, sample means are approximately normal when n ≥ 30 or when the population is already normal. Since reading speeds are normally distributed, sample means are normal for any sample size, including n = 25.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-07-10