I've seen this question few days before. Apparently, it is seemed little difficult. Although it is late, I think it is better to write now the solution
According to my opinion, in order to solve this problem, it is better to use the Central Limit Theorem.
: This theorem says
If a given population is distributed with an expectation of u-miu and a variance of sigma to square, the distribution of the average of a sample of big enough number of observation is normally distributed wuth an expectation of u – miu and a variance of sigma to square divided by n.
In this case the population is the teaching staff. The distribution is Normal. The expectation is 51 years and the variance is 8 years to square, namely 64
A group of nine-9 staff members is chosen at random.
According to the Central Limit Theorem, the expectation of the ages' average of the ages in the group is 51 years and the variance is the original variance divided by 9, namely 64/9. The standard deviation is 8/3 or 2.66 years. I think, that since the original staff's ages is normally distributed , then the number of 9 members is big enough so that the ages ' group averages is normally distributed too.
So X -The average of the ages of the chosen group is normally distributed
X~N(51,64/9) The standard deviation is 8/3 or 2.66 years
We want to know, what is the probability that this average is less than 52.9 years.
Let standardize X:
Let look at the normal distribution table. We found a probability of 0.56. This means that the probability that the age's average of the chosen group is less than 52.9 years is 0.56
If there are comments/questions, you can send a mail to [email protected]