Mathematics 2009 Jamb Examination

The mean of grouped data is given by:\[ \text{Mean} = \dfrac{\sum fx}{\sum f} \]The total frequency is:\[ 3 + 2 + 4 + 1 = 10 \]The sum of known products is:\[ (5 \times 3) + (8 \times 2) + (6 \times 4) = 15 + 16 + 24 = 55 \]Including k:\[ \dfrac{55 + k}{10} = 5.7 \]\[ 55 + k = 57 \]\[ k = 2 \]This satisfies the given mean.

First, determine the value of x using the mean.
The mean of the four numbers is:\[ \dfrac{x + 2x + (x + 1) + 3x}{4} = 2 \]\[ \dfrac{7x + 1}{4} = 2 \Rightarrow 7x + 1 = 8 \Rightarrow x = 1 \]The values are now 1, 2, 2 and 3, with mean 2.
Their absolute deviations from the mean are:
|1 − 2| = 1, |2 − 2| = 0, |2 − 2| = 0, |3 − 2| = 1.
The sum of deviations is 2.
The mean deviation is:\[ \dfrac{2}{4} = 0.5 \]This represents the average distance of the values from the mean.

The total number of people is 8, so the total number of ways of choosing any 3 people is:\[ \binom{8}{3} = 56 \]Delegations with only men are:\[ \binom{5}{3} = 10 \]Delegations with only women are:\[ \binom{3}{3} = 1 \]These cases do not satisfy the condition and must be excluded.
The required number of ways is:\[ 56 - (10 + 1) = 45 \]This ensures that each delegation contains at least one man and one woman.

Since only 3 chairs are available, the arrangement involves selecting and ordering 3 people out of 9.
This is a permutation problem given by:\[ {}^{9}P_{3} = 9 \times 8 \times 7 \]\[ = 504 \]The order matters because different seating positions produce different arrangements.

Passing the test means scoring 5 marks or above.
The number of students who passed is:\[ 0 + 4 + 5 + 2 + 1 = 12 \]The total number of students is:\[ 3 + 4 + 1 + 0 + 4 + 5 + 2 + 1 = 20 \]The probability of passing is therefore:\[ \dfrac{12}{20} = \dfrac{3}{5} \]This represents the proportion of students who met or exceeded the pass mark.