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WAEC 2013 · Paper 2 · Q11

The histogram shows the height of a group of students in a school.

144.5149.5154.5159.5164.5169.5174.5179.52468101214Height in cmNumber of students
  1. (a)

    Use the histogram to construct the frequency distribution table.

    Model answer
    Height (cm) 150–154 155–159 160–164 165–169 170–174 175–179
    Class boundaries 149.5–154.5 154.5–159.5 159.5–164.5 164.5–169.5 169.5–174.5 174.5–179.5
    Class mark xx 152 157 162 167 172 177
    Frequency ff 4 7 9 13 5 2
    fxfx 608 1099 1458 2171 860 354

    Σf=40\Sigma f = 40 and Σfx=6550\Sigma fx = 6550.

  2. (b)

    What percentage of the students have their heights between 159.5 cm159.5\text{ cm} and 164.5 cm164.5\text{ cm}?

  3. (c)

    Calculate the mean height.

Worked solution (try it first)

(a)

  1. The height of each bar is the frequency of the class between its two boundaries.
  2. The class mark is the middle of the class, for example 149.5+154.52=152\frac{149.5 + 154.5}{2} = 152.
  3. Height (cm) 150–154 155–159 160–164 165–169 170–174 175–179
    Class boundaries 149.5–154.5 154.5–159.5 159.5–164.5 164.5–169.5 169.5–174.5 174.5–179.5
    Class mark xx 152 157 162 167 172 177
    Frequency ff 4 7 9 13 5 2
    fxfx 608 1099 1458 2171 860 354
  4. Total frequency: 4+7+9+13+5+2=404 + 7 + 9 + 13 + 5 + 2 = 40 students.

(b)

  1. The bar from 159.5 to 164.5 has height 9, so 9 of the 40 students: 940×100%=22.5%\frac{9}{40} \times 100\% = 22.5\%.

(c)

  1. Add the fxfx row: Σfx=6550\Sigma fx = 6550.
  2. Mean =ΣfxΣf= \frac{\Sigma fx}{\Sigma f}
    =655040= \frac{6550}{40}
    =163.75= 163.75.
  3. The mean height is 163.75 cm163.75\text{ cm}.

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