Answer:
Step-by-step explanation:
Let x be the random variable representing the scores for the Algebra 2 CFE. Since they are normally distributed and the population mean and population standard deviation are known, we would apply the formula,
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
From the information given,
µ = 43
σ = 3.8
the probability that a test taker scores less than 47 is expressed as
P(x ≤ 47)
For x = 47
z = (47 - 43)/3.8 = 1.05
Looking at the normal distribution table, the probability corresponding to the area to the left of the z score is 0.85314
The percentage of test takers that would score lesser than you is
0.85314 × 100 = 85.314%
