Answer to Question #254688 in Statistics and Probability for Lisa

Question

(a)The table below show Business statistics I examination marks for a class of one hundred students in Mount Kenya university.

Marks

10-19

20-29

30-39

40-49

50-59

60-69

Number of students

4

8

a

22

48

b

The mean score of the students was 46.5

         (i) Show that the values of a and b are 12 and 6 respectively.                       (5marks)

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Marks & Number\ of \ students & C.F. \\ \hline 10-19 & 4 & 185 \\ \hdashline 20-29 & 8 & 12 \\ \hdashline 30-39 & a & 12+a \\ \hdashline 40-49 & 22 & 34+a \\ \hdashline 50-59 & 48 & 82+a \\ \hdashline 60-69 & b & 82+a+b \\ \hdashline \end{array}$$

The total frequency is $82+a+b$

$$82+a+b=100$$

$$b=18-a$$

The mean score of the students was 46.5.

$$14.5(4)+24.5(8)+34.5a+44.5(22)$$

$$+54.5(48)+64.5b=3849+34.5a+64.5b$$

$$mean=\dfrac{3849+34.5a+64.5b}{82+a+b}=46.5$$

$$3849+34.5a+64.5b=3813+46.5a+46.5b$$

$$18b+36=12a$$

$$3b=2a-6$$

Substitute

$$3(18-a)=2a-6$$

$$5a=60$$

$$a=12$$

$$b=6$$