Math Problem Statement

A population of values has a normal distribution with u = 141.5 and o = 95.2 . If a random sample of size n = 18 is selected,

Find the probability that a single randomly selected value is greater than 208.8. Round your answer to four decimals. P(X > 208.8) =

Find the probability that a sample of size n = 18 is randomly selected with a mean greater than 208.8. Round your answer to four decimals. P(M > 208.8) =

Solution

We are given a normally distributed population with:

  • Population mean μ=141.5\mu = 141.5
  • Population standard deviation σ=95.2\sigma = 95.2
  • A random sample of size n=18n = 18

Now, we are asked to find two probabilities.

1. Probability that a single randomly selected value is greater than 208.8:

We need to calculate P(X>208.8)P(X > 208.8).

For a single value from a normal distribution, we first standardize the value using the z-score formula: z=Xμσz = \frac{X - \mu}{\sigma} where:

  • X=208.8X = 208.8
  • μ=141.5\mu = 141.5
  • σ=95.2\sigma = 95.2

Let's calculate this z-score.

z=208.8141.595.2z = \frac{208.8 - 141.5}{95.2}

Next, we'll find the probability from the standard normal distribution table for this z-score, and subtract that from 1 to get P(X>208.8)P(X > 208.8).

2. Probability that a sample mean is greater than 208.8:

For the sample mean, we use the sampling distribution of the sample mean, which has a mean μ=141.5\mu = 141.5 and a standard deviation of σXˉ\sigma_{\bar{X}}, called the standard error of the mean: σXˉ=σn\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}

So, the standard error σXˉ\sigma_{\bar{X}} is: σXˉ=95.218\sigma_{\bar{X}} = \frac{95.2}{\sqrt{18}}

Then, we calculate the z-score for the sample mean: z=MμσXˉz = \frac{M - \mu}{\sigma_{\bar{X}}} where M=208.8M = 208.8, μ=141.5\mu = 141.5, and σXˉ\sigma_{\bar{X}} is what we calculated above.

Finally, we find the probability from the z-score table, and subtract that from 1 to get P(M>208.8)P(M > 208.8).

I'll calculate both these z-scores and probabilities for you now.Here are the results:

  1. The probability that a single randomly selected value is greater than 208.8 is: P(X>208.8)=0.2398P(X > 208.8) = 0.2398

  2. The probability that a sample of size n=18n = 18 is randomly selected with a mean greater than 208.8 is: P(M>208.8)=0.0014P(M > 208.8) = 0.0014

Would you like more details on these calculations or any additional clarifications?

Here are 5 related questions to further explore:

  1. How do you calculate the probability for a value less than a given number in a normal distribution?
  2. What happens to the probability if the sample size increases significantly?
  3. What is the effect of standard deviation on the probabilities for normal distributions?
  4. Can you explain why the probability for the sample mean is much smaller than for a single value?
  5. How can you use z-scores to compare different normal distributions?

Tip: As the sample size increases, the sampling distribution of the sample mean becomes narrower, which affects probabilities.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Probability
Z-scores
Sampling Distribution

Formulas

Z-score formula for single value: z = (X - μ) / σ
Standard error of the mean: σ_X̄ = σ / √n
Z-score formula for sample mean: z = (M - μ) / σ_X̄

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 11-12, College Level Statistics