Question

Suppose the scores on a test given to all juniors in a school district are normally distributed with a mean of 79 and a standard deviation of 7. find the probability that a randomly chosen junior has a score above 95.
the probability is _________.

Answer 1

Answer:

There is a 1.1% chance that a randomly chosen junior will have a score above 95.

Step-by-step explanation:

Calculate the z-score of 95 given a normal distribution with mean 79 and standard deviation of 7. The z-score is a probability (area under the distribution curve) of a value on a normalized random variable that has a distribution with mean 0 and standard deviation of 1. To get such a transformation, you need to subtract the mean and divide by the standard deviation, like this:

$z(x) = \frac{x-\mu}{\sigma}\\z(95)=\frac{95-79}{7}=2.29$

Based on this z-value, use z-score tables to look up the area under the curve. This will be the probability that randomly chosen values are lower (or higher, depending which table you are using) than this z-score. Be careful to apply the correct table. I found the following probability that a random z value is higher that the z-score 2.29: 0.011, or 1.1%. This means that there is a 1.1% chance that a randomly chosen junior will have a score above 95.