In a certainâ state, the recent average critical reading standardized test score was 514. Assume that the standard deviation is 50 and that standardized test scores are Normally distributed. Complete partsâ (a) andâ (b) below. Include a Normal curve for each part. Required:
a. What percentage of standardized test takers scored 550 or less?
b. What percentage of standardized test takers scored 524?

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!!!