Question

1. on a standardized exam, the scores are normally distributed with a mean of 300 and a standard deviation of 40. find the z - score of a person who scored 330 on the exam.

given:
μ = 300
σ = 40
x = 330
z=\frac{330 - 300}{40}=\frac{30}{40}=0.75
the z - score of the person who scored 330 on the exam is z = 0.75
a) what percent of the class score 350 and higher?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 350$, $\mu=300$, and $\sigma = 40$. So $z=\frac{350 - 300}{40}=\frac{50}{40}=1.25$.

Step2: Find the proportion in the standard normal table

We want to find $P(X\geq350)$, which is equivalent to $P(Z\geq1.25)$ in the standard - normal distribution. Since the total area under the standard - normal curve is 1, and the standard - normal table gives $P(Z < 1.25)=0.8944$. Then $P(Z\geq1.25)=1 - P(Z < 1.25)=1 - 0.8944 = 0.1056$.

Step3: Convert to percentage

To convert the proportion to a percentage, we multiply by 100. So the percentage is $0.1056\times100 = 10.56\%$.

Answer:

$10.56\%$