Central limit theorem

Question

A population of values has a normal distribution with mu = 78.9 and sigma = 48.5 . You intend to draw a random sample of size n = 65 Please show your answers as numbers accurate to 4 decimal places.

Find the probability that a single randomly selected value is greater than 69.3.

P(X > 69.3) =

Find the probability that a sample of size n = 65 is randomly selected with a mean greater than 69.3.

P( overline x > 69.3 )=

Answer 1

We can use the standard normal distribution to find the probabilities for this problem. To do this, we need to standardize the values using the formula:

z = (x - mu) / sigma

where x is the value, mu is the mean, sigma is the standard deviation, and z is the corresponding z-score from the standard normal distribution.

a) To find the probability that a single randomly selected value is greater than 69.3, we can standardize 69.3 as follows:

z = (69.3 - 78.9) / 48.5 = -0.1980

Using a standard normal distribution table or calculator, we can find the probability:

P(X > 69.3) = P(Z > -0.1980) ≈ 0.5796

Therefore, the probability that a single randomly selected value is greater than 69.3 is approximately 0.5796.

b) To find the probability that a sample of size n = 65 is randomly selected with a mean greater than 69.3, we can use the central limit theorem. Since the population is normally distributed, the sample means will also be normally distributed with mean mu = 78.9 and standard deviation sigma/sqrt(n) = 48.5/sqrt(65) ≈ 5.9967.

We can standardize the sample mean as follows:

z = (x̄ - mu) / (sigma / sqrt(n)) = (69.3 - 78.9) / (48.5 / sqrt(65)) ≈ -1.3287

Using a standard normal distribution table or calculator, we can find the probability:

P(overline x > 69.3) = P(Z > -1.3287) ≈ 0.9071

Therefore, the probability that a sample of size n = 65 is randomly selected with a mean greater than 69.3 is approximately 0.9071.