A population has a mean of 84 and a standard deviation of 12. A sample of 36 observations will be taken. The probability that the sample mean will be between 80.54 and 88.9 is

The probability that the sample mean will be between 80.54 and 88.9 is 0.951

Step-by-step explanation:

* Lets revise some definition to solve the problem

- The mean of the distribution of sample means is called M

- The standard deviation of the distribution of sample means is

called σM

- σM = σ/√n, where σ is the standard deviation and n is the sample size

- z-score = (M - μ) / σM, where μ is the mean of the population

* Lets solve the problem

∵ The sample size n = 36

∵ The sample mean M is between 80.54 and 88.9

∵ The mean of population μ = 84

∵ The standard deviation σ = 12

- Lets find σM to find z-score

∵ σM = σ/√n

∴ σM = 12/√36 = 12/6 = 2

- Lets find z-score

∵ z-score = (M - μ) / σM

∴ z-score = (80.54 - 84) / 2 = - 3.46/2 = - 1.73

∴ z-score = (88.9 - 84) / 2 = 4.9/2 = 2.45

- Use the normal distribution table to find the probability

∵ P (-1.73 < z < 2.45) = P (2.45) - P (-1.73)

∴ P (-1.73 < z < 2.45) = 0.99286 - 0.04182 = 0.95104

∴ P (-1.73 < z < 2.45) = 0.951

* The probability that the sample mean will be between 80.54 and 88.9

is 0.951