Math Problem Statement

A population of values has a normal distribution with mean μ = 151.6 and standard deviation σ = 21.5. You intend to draw a random sample of size n = 188. Find the probability that a single randomly selected value is greater than 154.7, and the probability that a sample of size n = 188 is randomly selected with a mean greater than 154.7.

Solution

Let's analyze and solve the problem step-by-step.

Problem Recap:

  1. A population is normally distributed with:
    • Mean (μ\mu) = 151.6
    • Standard deviation (σ\sigma) = 21.5
  2. You are tasked with:
    • Finding the probability P(X>154.7)P(X > 154.7) for a single randomly selected value.
    • Finding the probability P(M>154.7)P(M > 154.7), where MM is the sample mean from a sample of size n=188n = 188.

Step 1: Finding P(X>154.7)P(X > 154.7)

For a single value, use the standard normal distribution:

  1. Calculate the z-score: z=Xμσ=154.7151.621.5=3.121.50.1442z = \frac{X - \mu}{\sigma} = \frac{154.7 - 151.6}{21.5} = \frac{3.1}{21.5} \approx 0.1442

  2. Find the probability from the z-score:

    • Using a z-table or a statistical calculator, find the area to the left of z=0.1442z = 0.1442: P(Z<0.1442)0.5574P(Z < 0.1442) \approx 0.5574
    • The probability to the right is: P(X>154.7)=1P(Z<0.1442)=10.5574=0.4426P(X > 154.7) = 1 - P(Z < 0.1442) = 1 - 0.5574 = 0.4426

So, P(X>154.7)=0.4426P(X > 154.7) = 0.4426. (This matches the answer you entered.)


Step 2: Finding P(M>154.7)P(M > 154.7)

For the sample mean, adjust the standard deviation using the formula for the standard error: Standard Error (SE)=σn=21.518821.513.7111.567\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{21.5}{\sqrt{188}} \approx \frac{21.5}{13.711} \approx 1.567

  1. Calculate the z-score for the sample mean: z=MμSE=154.7151.61.567=3.11.5671.978z = \frac{M - \mu}{\text{SE}} = \frac{154.7 - 151.6}{1.567} = \frac{3.1}{1.567} \approx 1.978

  2. Find the probability from the z-score:

    • Using a z-table or a statistical calculator, find the area to the left of z=1.978z = 1.978: P(Z<1.978)0.9762P(Z < 1.978) \approx 0.9762
    • The probability to the right is: P(M>154.7)=1P(Z<1.978)=10.9762=0.0238P(M > 154.7) = 1 - P(Z < 1.978) = 1 - 0.9762 = 0.0238

So, P(M>154.7)=0.0238P(M > 154.7) = 0.0238. (This is close to the answer you entered, but your value might differ slightly due to rounding.)


Final Answers:

  1. P(X>154.7)=0.4426P(X > 154.7) = 0.4426 (correctly computed).
  2. P(M>154.7)=0.0238P(M > 154.7) = 0.0238.

Would you like a more detailed breakdown of any step or have further questions?


Related Questions:

  1. How do you calculate the standard error for a sample mean?
  2. Why do probabilities for sample means decrease as sample size increases?
  3. What is the importance of using z-scores in a normal distribution?
  4. How do rounding errors impact final probability calculations?
  5. What tools or calculators are most accurate for normal distribution probabilities?

Tip:

Always double-check whether you are working with single values or sample means, as the formulas and standard deviations differ.

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Math Problem Analysis

Mathematical Concepts

Probability
Normal Distribution
Z-Scores
Sampling Distribution

Formulas

z = (X - μ) / σ
z = (M - μ) / (σ / √n)

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate Statistics or Advanced High School