Question

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.

Answer 1

Answer:

The remaining two observations are 6 and 8

Step-by-step explanation:

Formula used:

$Mean=\frac{\Sigma{x}}{n}$

$variance=\frac{\Sigma(x-\bar{x})^2}{n}$

Let the remaining two observations be a and b

Given:

Mean = 8

$\frac{\Sigma{x}}{7}=8$

$\frac{2+4+10+12+14+a+b}{7}=8$

$\frac{42+a+b}{7}=8$

$42+a+b=56$

$a+b=14$............(1)

Variance =16

$\frac{\Sigma{(x-\bar{x})^2}}{7}=16$

$\Sigma{(x-\bar{x})^2}=112$

$\Sigma{(x-8)^2}=112$

$(2-8)^2+(4-8)^2+(10-8)^2+(12-8)^2+(14-8)^2+(a-8)^2+(b-8)^2=112$

$36+16+4+16+36+(a-8)^2+(b-8)^2=112$

$108+(a-8)^2+(b-8)^2=112$

$(a-8)^2+(14-a-8)^2=4$    (using(1))

$(a-8)^2+(6-a)^2=4$

$a^2+64-16a+36+a^2-12a=4$

$2a^2-28a+100=4$

$2a^2-28a+96=0$

Divide both sides by 2, we get

$a^2-14a+48=0$

$(a-6)(a-8)=0$

a = 6, 8

when a=6, b=8

when a=8, b=6

Answer 2

Answer:

6 & 8

Step-by-step explanation:

Let the other two observations be x and y.

Therefore,

our observations are 2,4,10,12,14,x, y.

Given Mean = 8

(2 + 4 + 10 + 12 + 14 + x + y)/7=8

42+x+y=7 x 8

x + y = 56 - 42

x+y = 14        .........(1)

Also,

Given Variance = 16

1/n∑(xi-X)=16

1/7[(2-8)² +(4-8)²+(10-8)²+(12-8)²+(14- 8)²+(x-8)²+(y- 8)²]= 16

1/7[(-6)² + (-4)² + (2)² + (4)² + (6)² + (x-8)² + (y - 8)²] = 16

1/7(36+16+4+16+36+x² +(8)²-2(8)x+ y²+(8)2-2(8)y] =16

[108+ x²+64-16x+y²+64 - 16y] = 16 x 7

[236+x²+y²-16y-16x ] =112

[236+x² + y²-16(x + y)] = 112

[236+ x²+y²-16(14)]=112

236 + x² + y²-224=112

x² + y² = 112-236 +224

x² + y² = 100         ..........(2)

From (1)

x+y 14

Squaring both sides

(x + y)² = 14

x² + y² + 2xy = 196

100 + 2xy = 196

2xy = 196 - 100

2xy = 96

xy= 1/2(96)

xy = 48

X=48/y

Putting (3) in (1)

x+y=14

48/y+y2=14

48 + y = 14

y²-14y+48 = 0

y²-6y - 8y +48 = 0

Y(y-6) -8(y-6)=0

(y-6)(y - 8) = 0

So,

y = 6 & y = 8

For y = 6

X=48/y

  = 48/6

  =8

Hence x = 8, y 6 are the remaining two observations

For y = 8

X=48/8

  =48/8

  =6

Hence, x=6, y=8 are the remaining two observations

Thus, remaining observations are 6 & 8