Answer:
a) Percentage of standardized test takers that scored 550 or less = 76.4%
b) Percentage of standardized test takers that scored 524 = 0.782%
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 514
Standard deviation = σ = 50
a) Percentage of standardized test takers scored 550 or less = P(x ≤ 550)
We first normalize or standardize 550
The standardized score for any is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (550 - 514)/50 = 0.72
To determine the required probability
P(x ≤ 550) = P(z ≤ 0.72)
We'll use data from the normal distribution table for these probabilities
P(x ≤ 550) = P(z ≤ 0.72) = 0.76424 = 76.424%
The normal curve for this question and the b part are sketched in the first attached image to this solution.
b) Percentage of standardized test takers that scored 524 = P(x = 524)
On standardizing,
z = (x - μ)/σ = (524 - 514)/50 = 0.20
For this part, since it's an exact probability, we will use the normal distribution formula
P(z = Z) = [1/(σ√2π)] × e^(-z²/2)
Since z = (x - μ)/σ
It can be written properly as presented in the second attached image to this question.
Putting x = 524 or z = 0.20 in this expression, we get
P(x = 524) = P(z = 0.20) = 0.0078208539 = 0.782%
Hope this Helps!!!
