Consider the \(60\) balls in the sack as the population. Let the
random variable \(X\) be the number marked on a ball randomly taken
out from this population. The distribution of \(X\), which is the
distribution of the population, is as the table below.
| \(X\) |
\(1\) |
\(2\) |
\(3\) |
Total |
| \(\mathrm{P}(X=x)\) |
\(\dfrac{1}{6}\) |
\(\dfrac{1}{3}\) |
\(\dfrac{1}{2}\) |
\(1\) |
Therefore the mean \(m\) and variance \(\sigma^2\) of the population
are
\(m=\mathrm{E}(X)=\dfrac{7}{3}\:\) and
\(\:\sigma^2=\mathrm{V}(X)=\fbox{\(\;(\alpha)\;\)}\).
Let \(\overline{X}\) be the mean of a random sample of size \(10\)
from this population. Then,
\(\mathrm{E}(\overline{X})=\dfrac{7}{3}\:\) and
\(\:\mathrm{V}(\overline{X})=\fbox{\(\;(\beta)\;\)}\).
Let \(X_n\) be the number marked on the \(n\:\!\)th ball taken out
from the sack. Then,
\(\displaystyle Y=\sum_{n\;\!=\;\!1}^{10}X_n=10\overline{X}\).
Therefore
\(\mathrm{E}(Y)=\dfrac{70}{3}\:\) and
\(\:\mathrm{V}(Y)=\fbox{\(\;(\gamma)\;\)}\).
Let \(p\), \(q\) and \(r\) be the correct numbers for \((\alpha)\),
\((\beta)\) and \((\gamma)\) respectively. What is the value of
\(p+q+r\)?