Respuesta :
Given:
μ = 112, population mean
σ = 25, population std. deviation
The randomly selected sample population of 300 (>30) is large enough for probability testing based on the normal distribution.
x = 115, random variable.
z-score:
x = (x-μ)/σ = (115 - 112)/25 = 0.12
Let xs = sample mean.
From standard tables, obtain
P(xs<115) = 0.548
Therefore
P(xs>115) = 1 - 0.548 = 0.452 = 45% (approx)
Answer: 45%
μ = 112, population mean
σ = 25, population std. deviation
The randomly selected sample population of 300 (>30) is large enough for probability testing based on the normal distribution.
x = 115, random variable.
z-score:
x = (x-μ)/σ = (115 - 112)/25 = 0.12
Let xs = sample mean.
From standard tables, obtain
P(xs<115) = 0.548
Therefore
P(xs>115) = 1 - 0.548 = 0.452 = 45% (approx)
Answer: 45%
Answer:
The correct asnwer is 0.019
Step-by-step explanation:
I Just did it, 0.45 is incorrect
