Question

Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.542.54 and a standard deviation of 0.450.45. Using the empirical rule, what percentage of the students have grade point averages that are greater than 3.443.44

Answer 1

Answer:

$z= \frac{3.44-2.54}{0.45}=2$

$P(X>3.44) = P(Z>2) = 0.025$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the grade points avergae of a population, and for this case we know the following properties

Where $\mu=2.54$ and $\sigma=0.45$

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

So we can find the z score for the value of X=3.44 in order to see how many deviations above or belowe we are from the mean like this:

$z= \frac{3.44-2.54}{0.45}=2$

So the value of 3.44 is 2 deviations above from the mean, so then we know that the percentage between two deviations from the mean is 95% and on each tail we need to have (100-95)/2 = 2.5% , because the distribution is symmetrical, so based on this we can conclude that:

$P(X>3.44) = P(Z>2) = 0.025$