Math Problem Statement

A population of values has a normal distribution with μ = 53.9 and σ = 6.1. a) Find the probability that a single randomly selected value is between 51.2 and 56.8. b) Find the probability that a randomly selected sample of size n = 44 has a mean between 51.2 and 56.8.

Solution

Let's solve the questions step-by-step.

Given Information:

  • The population has a normal distribution with a mean (μ\mu) of 53.9 and a standard deviation (σ\sigma) of 6.1.

Part (a):

We are asked to find the probability that a single randomly selected value falls between 51.2 and 56.8. This is calculated as:

P(51.2<X<56.8)P(51.2 < X < 56.8)

To solve this, we need to calculate the Z-scores for 51.2 and 56.8 using the formula:

Z=XμσZ = \frac{X - \mu}{\sigma}

  1. For X=51.2X = 51.2:

Z51.2=51.253.96.1=2.76.10.4426Z_{51.2} = \frac{51.2 - 53.9}{6.1} = \frac{-2.7}{6.1} \approx -0.4426

  1. For X=56.8X = 56.8:

Z56.8=56.853.96.1=2.96.10.4754Z_{56.8} = \frac{56.8 - 53.9}{6.1} = \frac{2.9}{6.1} \approx 0.4754

Next, we find the probability corresponding to these Z-scores from the standard normal distribution table or using a calculator.

  • P(Z=0.4426)0.3298P(Z = -0.4426) \approx 0.3298
  • P(Z=0.4754)0.6823P(Z = 0.4754) \approx 0.6823

Finally, the probability is:

P(51.2<X<56.8)=P(Z=0.4754)P(Z=0.4426)0.68230.3298=0.3525P(51.2 < X < 56.8) = P(Z = 0.4754) - P(Z = -0.4426) \approx 0.6823 - 0.3298 = 0.3525

So, the probability for part (a) is approximately 0.3525.

Part (b):

For part (b), we are asked to find the probability that a randomly selected sample of size n=44n = 44 has a mean between 51.2 and 56.8. This requires the standard error of the mean, which is calculated as:

Standard Error=σn=6.1446.16.63320.9194\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{6.1}{\sqrt{44}} \approx \frac{6.1}{6.6332} \approx 0.9194

Now, we calculate the Z-scores for the sample mean using this standard error:

  1. For M=51.2M = 51.2:

Z51.2=51.253.90.9194=2.70.91942.9362Z_{51.2} = \frac{51.2 - 53.9}{0.9194} = \frac{-2.7}{0.9194} \approx -2.9362

  1. For M=56.8M = 56.8:

Z56.8=56.853.90.9194=2.90.91943.1547Z_{56.8} = \frac{56.8 - 53.9}{0.9194} = \frac{2.9}{0.9194} \approx 3.1547

We now find the probabilities corresponding to these Z-scores:

  • P(Z=2.9362)0.0017P(Z = -2.9362) \approx 0.0017
  • P(Z=3.1547)0.9992P(Z = 3.1547) \approx 0.9992

So, the probability is:

P(51.2<M<56.8)=P(Z=3.1547)P(Z=2.9362)0.99920.0017=0.9975P(51.2 < M < 56.8) = P(Z = 3.1547) - P(Z = -2.9362) \approx 0.9992 - 0.0017 = 0.9975

Thus, the probability for part (b) is approximately 0.9975.


Let me know if you'd like more details or have any questions!

Related questions:

  1. What are Z-scores, and how are they used in probability calculations?
  2. How does sample size affect the standard error in normal distributions?
  3. How can we calculate probabilities for non-standard normal distributions?
  4. What is the central limit theorem and how does it relate to sample means?
  5. How do we interpret probabilities in the context of normal distributions?

Tip:

Always check if you're working with a population or sample, as this affects whether you need to use the standard error or population standard deviation when calculating probabilities!

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-scores
Standard Error

Formulas

Z = (X - μ) / σ
Standard Error = σ / √n

Theorems

Central Limit Theorem

Suitable Grade Level

College/University