Let's denote the five observations as
x1,x2,x3,x4, and
x5. We know the following:
1. The mean of the observations is 4 . Therefore, we can write:
‌=4Multiplying both sides by 5 , we get:
x1+x2+x3+x4+x5=202. We are given three of these observations: 1,2 , and 6 . Let's assume:
x1=1,x2=2,x3=6Substituting these values into the sum equation:
1+2+6+x4+x5=20 Simplifying this, we get:
‌9+x4+x5=20‌x4+x5=11Now, let's use the given variance to determine the values of
x4 and
x5.
3. The variance of the observations is 5.2. The variance formula for a sample is given by:
‌ Variance ‌=‌‌(xi−µ)2 where
µ is the mean, which is 4 in this case.
So, we need to calculate:
‌(xi−4)2=5×‌ Variance ‌‌(xi−4)2=5×5.2=26 We can sum the squares of the deviations for the given observations:
‌(1−4)2+(2−4)2+(6−4)2+(x4−4)2+(x5−4)2=26
‌(3)2+(2)2+(2)2+(x4−4)2+(x5−4)2=26
‌9+4+4+(x4−4)2+(x5−4)2=26‌17+(x4−4)2+(x5−4)2=26‌(x4−4)2+(x5−4)2=9 We now need to solve these two equations:
‌x4+x5=11‌(x4−4)2+(x5−4)2=9We can test the options given:
Option A:
x4=4,x5=7(4−4)2+(7−4)2=0+9=9 Option A:
x4=4,x5=7(4−4)2+(7−4)2=0+9=9This is a valid solution.
Option B:
x4=2,x5=9(2−4)2+(9−4)2=4+25=29This is not a valid solution.
Option C:
x4=5,x5=6(5−4)2+(6−4)2=1+4=5This is not a valid solution.
Option D:
x4=2,x5=9(2−4)2+(9−4)2=4+25=29This is also not a valid solution.
Therefore, the correct answer is Option A:
x4=4 and
x5=7.
This is a valid solution.