Math Problem Statement

A population of values has a normal distribution with mean (μ) = 27.4 and standard deviation (σ) = 96.8. You intend to draw a random sample of size n = 168. Find the probability that a single randomly selected value is between 8.7 and 28.1. Also, find the probability that a sample of size n = 168 is randomly selected with a mean between 8.7 and 28.1.

Solution

Let's break down both parts of this problem step by step.

1. Probability for a single randomly selected value:

We are asked to find the probability that a single randomly selected value XX falls between 8.7 and 28.1, given that the population has a normal distribution with mean μ=27.4\mu = 27.4 and standard deviation σ=96.8\sigma = 96.8.

We will use the z-score formula to transform these values into standard normal distribution values:

z=Xμσz = \frac{X - \mu}{\sigma}

  • For X=8.7X = 8.7: z1=8.727.496.8=18.796.80.1932z_1 = \frac{8.7 - 27.4}{96.8} = \frac{-18.7}{96.8} \approx -0.1932

  • For X=28.1X = 28.1: z2=28.127.496.8=0.796.80.0072z_2 = \frac{28.1 - 27.4}{96.8} = \frac{0.7}{96.8} \approx 0.0072

Now, using a standard normal table or a calculator:

  • The cumulative probability corresponding to z1=0.1932z_1 = -0.1932 is approximately 0.4235.
  • The cumulative probability corresponding to z2=0.0072z_2 = 0.0072 is approximately 0.5029.

Thus, the probability that XX is between 8.7 and 28.1 is: P(8.7<X<28.1)=0.50290.4235=0.0794P(8.7 < X < 28.1) = 0.5029 - 0.4235 = 0.0794

2. Probability for a sample mean:

Now, we are finding the probability that a sample of size n=168n = 168 has a mean between 8.7 and 28.1. For this, we use the sampling distribution of the sample mean. The standard deviation of the sample mean (standard error) is given by:

σXˉ=σn=96.81687.467\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{96.8}{\sqrt{168}} \approx 7.467

We again use the z-score formula, but this time for the sample mean:

z=MμσXˉz = \frac{M - \mu}{\sigma_{\bar{X}}}

  • For M=8.7M = 8.7: z1=8.727.47.467=18.77.4672.504z_1 = \frac{8.7 - 27.4}{7.467} = \frac{-18.7}{7.467} \approx -2.504

  • For M=28.1M = 28.1: z2=28.127.47.467=0.77.4670.094z_2 = \frac{28.1 - 27.4}{7.467} = \frac{0.7}{7.467} \approx 0.094

Now, using a standard normal table or a calculator:

  • The cumulative probability corresponding to z1=2.504z_1 = -2.504 is approximately 0.0062.
  • The cumulative probability corresponding to z2=0.094z_2 = 0.094 is approximately 0.5374.

Thus, the probability that the sample mean is between 8.7 and 28.1 is: P(8.7<M<28.1)=0.53740.0062=0.5312P(8.7 < M < 28.1) = 0.5374 - 0.0062 = 0.5312

Final Answers:

  1. P(8.7<X<28.1)=0.0794P(8.7 < X < 28.1) = 0.0794
  2. P(8.7<M<28.1)=0.5312P(8.7 < M < 28.1) = 0.5312

Would you like more clarification on any of the steps?

Here are 5 related questions:

  1. How would the probability change if the sample size increased to 200?
  2. What is the probability that a single randomly selected value is less than 27.4?
  3. How do you find the z-scores for values far outside the mean, like 50 or 100?
  4. What assumptions must hold for the sample mean to follow a normal distribution?
  5. How would the probabilities change if the standard deviation were 50 instead of 96.8?

Tip: When dealing with large sample sizes, the sampling distribution of the sample mean has less variability than the population, which is why the sample mean often results in higher probabilities compared to individual values.

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Math Problem Analysis

Mathematical Concepts

Probability
Normal Distribution
Z-Score
Sampling Distribution

Formulas

z = (X - μ) / σ
σ_X̄ = σ / √n
P(a < X < b) = Φ(z2) - Φ(z1)

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate (Statistics or Probability Course)