Answer:
![P(z>-1.768)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E-1.768%29%3D1-P%28z%3C-1.768%29%3D1-0.03853%3D0.96147)
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:
Let Y the scores for the univerisity B, and we know that:
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)](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%3D625%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B10%7D%7B%5Csqrt%7B2%7D%7D%3D7.07%29)
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)](https://tex.z-dn.net/?f=%5Cbar%20Y%20%5Csim%20N%28%5Cmu%3D600%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B12.25%7D%7B%5Csqrt%7B3%7D%7D%3D7.07%29)
Since both sample means are normally distributed then the difference
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)](https://tex.z-dn.net/?f=Z%3D%20%5Cbar%20X%20-%5Cbar%20Y%20%5Csim%20N%28%5Cmu_Z%3D625-600%3D25%2C%20%5Csigma_z%3D%20%5Csqrt%7B100%2B100%7D%3D14.14%29)
And we want this probability:
![P(Z>0)](https://tex.z-dn.net/?f=%20P%28Z%3E0%29)
And we can use the z score given by:
![Z= \frac{z -\mu_z}{\sigma_z}](https://tex.z-dn.net/?f=%20Z%3D%20%5Cfrac%7Bz%20-%5Cmu_z%7D%7B%5Csigma_z%7D)
And if we replace we got :
![Z= \frac{0-25}{14.14}=-1.768](https://tex.z-dn.net/?f=%20Z%3D%20%5Cfrac%7B0-25%7D%7B14.14%7D%3D-1.768)
And if we find the probability using the normla standard table or excel we got:
![P(z>1.768) =1-P(Z](https://tex.z-dn.net/?f=%20P%28z%3E1.768%29%20%3D1-P%28Z%3C-1.768%29%20%3D1-0.03853%3D0.96147)