Answer:
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
