Question

an instructor in a large lecture class found, at the end of the semester, that the total point distribution in his class was approximately normal, with a mean of 500 and a standard deviation of 80. about what percent did score above 590?
a. 0.1303
b. 0.8697
c. 0.5
d. not possible, the sample standard deviation is not given

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x = 590$, $\mu=500$, and $\sigma = 80$. So, $z=\frac{590 - 500}{80}=\frac{90}{80}=1.125$.

Step2: Find the proportion

We use the standard normal distribution table (or z - table). The table gives the proportion of values to the left of a given z - score. Looking up $z = 1.125$ in the z - table (or using a calculator with a normal - distribution function), the proportion of values to the left of $z = 1.125$ is approximately $0.8697$. The proportion of values above $z = 1.125$ is $1-0.8697 = 0.1303$.

Answer:

a. 0.1303