Central Limit Theorem

Question

A population of values has a normal distribution with  

μ=22.4 and σ =12.9 . You intend to draw a random sample of size N=187.

 Please show your answers as numbers accurate to 4 decimal places.

Find the probability that a single randomly selected value is greater than 23.2.

P(X > 23.2) = 

Find the probability that a sample of size �=187 is randomly selected with a mean greater than 23.2.

P(x¯ > 23.2) =

Answer 1

To solve this problem, we will use the properties of the normal distribution and the central limit theorem.

The probability that a single randomly selected value is greater than 23.2 can be found using the standard normal distribution:

Z = (X - μ) / σ

where X is the random variable, μ is the mean, and σ is the standard deviation.

Substituting the given values, we get:

Z = (23.2 - 22.4) / 12.9 = 0.06202

Using a standard normal table or calculator, we can find the probability that Z is greater than 0.06202:

P(Z > 0.06202) = 0.4756

Therefore, the probability that a single randomly selected value is greater than 23.2 is 0.4756.

The probability that a sample of size N = 187 is randomly selected with a mean greater than 23.2 can be found using the central limit theorem. Since N is large (greater than 30) and the population is normally distributed, the sample mean will also be normally distributed with mean μ and standard deviation σ/sqrt(N).

Substituting the given values, we get:

μ = 22.4 σ = 12.9 N = 187

The standard deviation of the sample mean is:

σ_x̄ = σ / sqrt(N) = 12.9 / sqrt(187) = 0.9438

The z-score corresponding to a sample mean of 23.2 is:

Z = (x̄ - μ) / (σ_x̄) = (23.2 - 22.4) / 0.9438 = 0.8435

Using a standard normal table or calculator, we can find the probability that Z is greater than 0.8435:

P(Z > 0.8435) = 0.1995

Therefore, the probability that a sample of size N = 187 is randomly selected with a mean greater than 23.2 is 0.1995.