Math Problem Statement

Assume that the scores on a placement exam are normally distributed with a mean of 70.4 and standard deviation of 8.34. If 8 scores are chosen at random, what is the probability that the sample mean will fall between 66 and 78?

Round your answer to 4 decimal places.

You Answered 0.9268 Correct Answer 0.9272

Solution

To solve this problem, we need to calculate the probability that the sample mean of 8 randomly chosen scores from a normally distributed population will fall between 66 and 78. Given:

  • Population mean (μ\mu) = 70.4
  • Population standard deviation (σ\sigma) = 8.34
  • Sample size (nn) = 8
  • Sample mean interval: 66 to 78

Steps:

  1. Determine the standard error (SE): The standard error of the sample mean is calculated by: SE=σnSE = \frac{\sigma}{\sqrt{n}} Substituting the given values: SE=8.3482.9496SE = \frac{8.34}{\sqrt{8}} \approx 2.9496

  2. Calculate the z-scores: Next, we convert the sample mean values (66 and 78) to z-scores using the formula: z=XμSEz = \frac{X - \mu}{SE}

    For 66: z1=6670.42.94961.4917z_1 = \frac{66 - 70.4}{2.9496} \approx -1.4917

    For 78: z2=7870.42.94962.5771z_2 = \frac{78 - 70.4}{2.9496} \approx 2.5771

  3. Find the probability for each z-score: Using a z-table or a calculator to find the cumulative probability:

    • The cumulative probability for z11.4917z_1 \approx -1.4917 is about 0.0682.
    • The cumulative probability for z22.5771z_2 \approx 2.5771 is about 0.9950.
  4. Find the probability that the sample mean falls between 66 and 78: The probability is the difference between the two cumulative probabilities: P(66X78)=P(z2)P(z1)=0.99500.0682=0.9268P(66 \leq X \leq 78) = P(z_2) - P(z_1) = 0.9950 - 0.0682 = 0.9268

So, the correct probability should be 0.9268. However, the answer you provided was incorrect because of rounding. The correct cumulative values from a more precise table or calculator would give:

P(66X78)=0.9272P(66 \leq X \leq 78) = 0.9272

This final value rounds to 0.9272 when using a more accurate z-distribution table or tool.

Would you like further details on any step?

Related Questions:

  1. How does the sample size affect the standard error?
  2. What happens to the probability if the population standard deviation changes?
  3. Can the z-score formula be used for any sample size?
  4. Why do we assume a normal distribution in this scenario?
  5. How does rounding affect the final probability result?

Tip: Always double-check the precision of z-tables or software tools when rounding, as even small differences can affect your final answer.

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Sampling Distribution

Formulas

Standard error (SE) = σ / √n
Z-score formula: z = (X - μ) / SE
P(X1 ≤ X ≤ X2) = P(z2) - P(z1)

Theorems

Central Limit Theorem
Empirical Rule

Suitable Grade Level

Grades 10-12