Question

for all of the normal distribution percentages (probabilities). you need to use the methods shown in class, which requires using a standard normal probability table. if you use technology to do these calculations for you, there might be rounding issues which will mark your answers as incorrect.
if you need a copy of the table, you can find it on this website table a - standard normal probabilities
the scores on a statistics exam had an approximately normal distribution, with a mean of 74 and standard deviation of 4.8. if a single student is chosen at random, what is the probability their score is less than 73?
question help: message instructor

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 73\), \(\mu=74\), and \(\sigma = 4.8\).

$$z=\frac{73 - 74}{4.8}=\frac{-1}{4.8}\approx - 0.21$$

Step2: Find the probability using the standard normal table

Looking up the z - score of \(-0.21\) in the standard normal table (Table A - Standard Normal Probabilities). The table gives the cumulative probability \(P(Z\lt z)\).
For \(z=-0.21\), the cumulative probability \(P(Z\lt - 0.21)\) is \(0.4168\)

Answer:

The probability that a student's score is less than \(73\) is approximately \(0.4168\)