Question

Suppose that on a certain examination in advanced mathematics, students from university A achieve scores normally distributed with a mean of 625 and a variance of 100, and students from university B achieve scores 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 from university B? Hint: Determine the distribution of the difference between the two averages

Short answer

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 from university B is 0.9938.

Step-by-step solution

Given information

Let\(\bar X\)the average scores of two students from university A be the average of three students from university B.

It is given that scores of students from university A are normally distributed with a mean of 625 and a variance of 100.i.e \(\bar X \sim N\left( {625,100} \right)\).

Also,university B students' scores are normally distributed with a mean of 600 and a variance of 150.i.e.,\(\bar Y \sim N\left( {600,150} \right)\)

Compute the Probability

Since\(\bar X\)it has a normal distribution with mean =625 and variance =\(\frac{{100}}{2}\)

i.e.,\(\bar X\)has a normal distribution with \(E\left( {\bar X} \right) = 625\)and\(V\left( {\bar X} \right) = 50\).

Also,\[\bar Y\]has a normal distribution with mean =600 and variance =\(\frac{{150}}{3}\)

i.e.,\[\bar Y\]has a normal distribution with \(E\left( {\bar Y} \right) = 600\)and\(V\left( {\bar Y} \right) = 50\).

Therefore, by the property of Normal distribution, if X and Y are independent normal variates, then\(E\left( {X - Y} \right) = E\left( X \right) - E\left( Y \right)\)and\(V\left( {X - Y} \right) = V\left( X \right) + V\left( Y \right)\)

It implies that \(E\left( {\bar X - \bar Y} \right) = E\left( {\bar X} \right) - E\left( {\bar Y} \right)\) and \(V\left( {\bar X - \bar Y} \right) = V\left( {\bar X} \right) - V\left( {\bar Y} \right)\)

Therefore,

\(\begin{array}{c}E\left( {\bar X - \bar Y} \right) = 625 - 600\\ = 25\end{array}\)

\[\begin{array}{c}V\left( {\bar X - \bar Y} \right) = 50 + 50\\ = 100\end{array}\]