Question

The sample mean and sample variance of five data values are, respectively 13.6 and 25.8. If three of the data values are 7, 13 and 20, what are the other two data values? ________, ________.

Answer 1

Answer:

19 and 9

Explanation:

by using the formulae of mean and variance

Answer 2

Answer:

The other two values are $11,17.$

Explanation:

Let the two unknown values be $x$ and $y$.

It is given that the arithmetic mean is $13.6$.

Arithmetic mean :

$AM =$ ∑$\frac{x_{i} }{n}$

$= \frac{7+13+20+x+y}{5}$ $=13.6$

$7+13+20+x+y=68$

$x+y=28$     --- Let this be (i)

Sample variance $=$ ∑$\frac{(x_{i}-AM)^{2}}{n-1}$ $=25.8$

$25.8 = \frac{(7-13.6)^{2} +(13-13.6)^{2} +(20-13.6)^{2} +(x-13.6)^{2} +(y-13.6)^{2} }{4} \\\\103.2 = 43.56+0.36+40.96+x^{2} +(13.6)^{2} -2x(13.6)+y^{2} +(13.6)^{2} -2y(13.6)$

From this we get :

$x^{2} +y^{2} =410$    --- Let this be (ii)

Now, taking square of (i) :

$x^{2} +y^{2} +2xy = 28^{2} \\x^{2} +y^{2} +2xy = 784$

Substituting values from (ii) :

$2xy=374$

Now,

$x^{2} +y^{2}-2xy=410-374\\ = 36$

$(x-y)^{2} =36$

$x-y=$±$6$      --- Let this be (iii)

From (i) and (iii):

When taking $x-y=6,$

$2x=34\\x=17\\y=11$

When taking $x-y=-6,$

$x=11\\y=17$

Hence the two missing values are $11,17.$