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WAEC 2009 · Paper 2 · Q3

The table shows the number of children per family in a community.

No. of children 0 1 2 3 4 5
No. of families 3 5 7 4 3 2
  1. (a)

    Find the: (i) mode; (ii) third quartile; (iii) probability that a family has at least 2 children.

    Separate values with commas, e.g. 3, −2

  2. (b)

    If a pie chart were to be drawn for the data, what would be the sectoral angle representing families with one child?

Worked solution (try it first)

(a)(i)

  1. The mode is the number of children with the highest frequency: 7 families have 2 children, so the mode is 2.

(ii)

  1. Total number of families: 3+5+7+4+3+2=243 + 5 + 7 + 4 + 3 + 2 = 24.
  2. The third quartile is at position 34×24=18\frac34 \times 24 = 18 in the ordered list.
  3. Running totals: 3 (0 children), 8 (1 child), 15 (2 children), 19 (3 children).
  4. The 16th to 19th families have 3 children, so the 18th has 3.
  5. The third quartile is 3 children.

(iii)

  1. Families with at least 2 children: 7+4+3+2=167 + 4 + 3 + 2 = 16.
  2. Probability =1624=23= \frac{16}{24} = \frac23.

(b)

  1. Angle for one child =524×360∘= \frac{5}{24} \times 360^\circ
    =75∘= 75^\circ.

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