## Elementary Statistics (12th Edition)

Here, n=12, K=10, N=40. In a hypergeometric distribution the mean can be counted by: $n \cdot \frac{K}{N}=12 \cdot \frac{10}{40}=3.$ In a hypergeometric distribution the variance can be counted by: $n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1}=12 \cdot \frac{10}{40} \cdot \frac{30}{40} \cdot \frac{28}{39}=1.615.$ The standard deviation is the square root of the variance: $\sqrt {1.615}=1.27.$