Math Problem Statement

Again following on from the previous question, what is the probability that when we take a random sample of n

78 adults, the sample mean would be either greater than 7.1187 or less than 6.8813?

Solution

To find the probability that the sample mean is either greater than 7.1187 or less than 6.8813, we'll use the concept of the sampling distribution of the sample mean.

Step 1: Determine the parameters of the sampling distribution

The sampling distribution of the sample mean Xˉ\bar{X} follows a normal distribution: XˉN(μ,σn)\bar{X} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) where:

  • μ\mu is the population mean,
  • σ\sigma is the population standard deviation,
  • nn is the sample size.

Given:

  • n=78n = 78
  • Population mean μ=7\mu = 7 (assumed from context, if not provided)
  • Population standard deviation σ=0.5\sigma = 0.5 (assumed from context, if not provided)

The standard error (SE) of the sample mean is: SE=σn=0.5780.0566\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{78}} \approx 0.0566

Step 2: Standardize the values

To find the probability, we convert the sample means to zz-scores using the formula: z=XˉμSEz = \frac{\bar{X} - \mu}{\text{SE}}

For Xˉ=7.1187\bar{X} = 7.1187: z=7.118770.05662.10z = \frac{7.1187 - 7}{0.0566} \approx 2.10

For Xˉ=6.8813\bar{X} = 6.8813: z=6.881370.05662.10z = \frac{6.8813 - 7}{0.0566} \approx -2.10

Step 3: Find the corresponding probabilities

Using standard normal distribution tables or a calculator:

  • P(Z>2.10)P(Z > 2.10) is the probability that the zz-score is greater than 2.10.
  • P(Z<2.10)P(Z < -2.10) is the probability that the zz-score is less than -2.10.

Since the normal distribution is symmetric, we have: P(Z>2.10)=P(Z<2.10)0.0179P(Z > 2.10) = P(Z < -2.10) \approx 0.0179

Step 4: Calculate the total probability

The total probability that the sample mean is either greater than 7.1187 or less than 6.8813 is the sum of these two probabilities: P(Xˉ>7.1187 or Xˉ<6.8813)=P(Z>2.10)+P(Z<2.10)0.0179+0.0179=0.0358P(\bar{X} > 7.1187 \text{ or } \bar{X} < 6.8813) = P(Z > 2.10) + P(Z < -2.10) \approx 0.0179 + 0.0179 = 0.0358

Final Answer:

The probability that the sample mean is either greater than 7.1187 or less than 6.8813 is approximately 0.0358.

Would you like more details or have any questions?

Here are some related questions you might be interested in:

  1. How does the sample size affect the probability of the sample mean being within a certain range?
  2. What assumptions are made about the population distribution when using the Central Limit Theorem?
  3. How would the probability change if the sample size were doubled?
  4. What is the difference between the probability of a single observation being within a range and the sample mean being within that range?
  5. How can we calculate probabilities for non-normal population distributions?

Tip: When dealing with sample means, increasing the sample size decreases the standard error, which narrows the distribution of the sample mean and affects the probabilities of extreme values.

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Math Problem Analysis

Mathematical Concepts

Probability
Sampling Distribution
Normal Distribution
Z-score

Formulas

Sampling distribution of the sample mean
Standard error formula
Z-score formula

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate