Question

Suppose that on a certain examination in advanced mathematics, students from univer sity A achieve scores that are normally distributed with a mean of 625 and a variance of 100, and students from university B achieves scores which are normally distributed with a mean of 600 and a variance of 150. If two students from university A and three students from university B take this examination, what is the probability that the average of the scores of the two students from university A will be greater than the average of the scores of the three students form university B? Hint: Determine the distribution of the difference between the two averages.

Answer 1

Answer:

$P(z>-1.768)=1-P(z<-1.768)=1-0.03853=0.96147$

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 scores for the univerisity A, and we know that:

$X \sim N (\mu = 625,\sigma =\sqrt{100}= 10)$

Let Y the scores for the univerisity B, and we know that:

$Y \sim N (\mu = 600,\sigma =\sqrt{150}= 12.25)$

We select a sample size of size n=2, and since the distirbution for X is normal then the distribution for the sample mean would be given by:

$\bar X \sim N(\mu=625, \frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{2}}=7.07)$

And for the univeristy B we select a sample of n=3

$\bar Y \sim N(\mu=600, \frac{\sigma}{\sqrt{n}}=\frac{12.25}{\sqrt{3}}=7.07)$

Since both sample means are normally distributed then the difference $Z= \bar X- \bar Y$ is also normal distributed with the following parameters:

$Z= \bar X -\bar Y \sim N(\mu_Z=625-600=25, \sigma_z= \sqrt{100+100}=14.14)$

And we want this probability:

$P(Z>0)$

And we can use the z score given by:

$Z= \frac{z -\mu_z}{\sigma_z}$

And if we replace we got :

$Z= \frac{0-25}{14.14}=-1.768$

And if we find the probability using the normla standard table or excel we got:

$P(z>1.768) =1-P(Z<-1.768) =1-0.03853=0.96147$