The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6, then the other two are:
4 and 7
Let the other two numbers be
α and β.
¯x=4,N=5 and ∑(x−¯x)2N=5.2⇒ ∑(x−¯x)2=(5.2)5∴ ∑(x−¯x)2=26∴ (1−4)2+(2−4)2+(6−4)2+(α−4)2+(β−4)2=26∴ (α−4)2+(β−4)2=9Also,1+2+6+α+β5=4∴ α+β=20−9=11Clearly 4,7 only satisfy the above equation in α,β.Hence required numbers are 4,7.