Question

The following data set has a mean of 14,7 and a variance of 10,01. 18,11,12,a,16,11,19,14,b,13
Compute the value of a and b.

Answer 1

Mean is the average of data set.

in this case:

mean=(18+11+12+a+16+11+19+14+b+14)/number of elements in a set= (114+a+b)/10

we know that mean=14.7, so

$$(114+a+b)/10=14.7$$

Multiplying both sides by 10:

$$114+a+b=147\implies a+b=33$$

The variance ( σ²) is a measure of how far each value in the data set is from the mean. Here is how it is defined:

1. Subtract the mean from each value in the data. This gives you a measure of the distance of each value from the mean.
2. Square each of these distances (so that they are all positive values), and add all of the squares together.
3. Divide the sum of the squares by the number of values in the data set.

In this case:

$σ^2=((18-14.7)^2+(11-14.7)^2+(12-14.7)^2+(a-14.7)^2+(16-14.7)^2+(11-14.7)^2+(19-14.7)^2+(14-14.7)^2+(b-14.7)^2+(13-14.7)^2)/10=10.01$

$variance( σ2)= 10.89+13.69+7.29+(a-14.7)^2+1.69+13.69+18.49+0.49+(b-14.7)^2+2.89=10*10.01=100.1\implies$

$(a-14.7)^2+(b-14.7)^2=100.1-69.12=30.98$

$(a^2-2*14.7*a+14.7^2)+(b^2-2*14.7*b+14.7^2)=30.98$

Substituting $a+b=33\implies a=33-b$ ,

$(33-b)^2-2*14.7*(33-b)+14.7^2+b^2-2*14.7*b+14.7^2=30.98$

$1089-2*33*b+b^2-2∗14.7∗(33−b)+14.7^2 +b^2 −2∗14.7∗b+14.7^2 =30.98$

$2*b^2-66b+550.98=30.98$

$b^2-33b+260=0$

$b^2-13b-20b+260=b(b-13)-20(b-13)=(b-13)(b-20)=0\implies b_1=13; b_2=20;$

substituting these values into $a+b=33, a_1=33-13=20 ; a_2=33-20=13;$

which means a and b take values 13 and 20.