Question

histograms: the maths GCSE test scores of 280 students are shown in the histogram below

Answer 1

Answer:

Median = 116

Answer 2

Quartiles and median can be calculated from histograms.

• The median is 116
• The lower quartile is 102

From the histogram, we have the following frequency table:

$\left[\begin{array}{ccc}Score&Frequency&Cumulative\\80 - 90&2.6&2.6&90 - 100&3.4&6&100 - 110&5&11&110 - 120& 5 & 16&120 - 130&3&19&130 - 140& 3 &22 &140 - 150&1.4&23.4& 150 - 160& 1.4 &24.8 & 160 - 170 & 1.4 &26.2 & 170 - 180 & 0.6 & 26.8& 180 - 190 & 0.6 & 27.4 &190 - 200&0.6&28\end{array}\right]$

(a) The median

Calculate the median class as follows

$\mathbf{Q_2 = \frac{1}{2}(n)}$

This gives

$\mathbf{Q_2 = \frac{1}{2}(28)th}$

$\mathbf{Q_2 =14th}$

The 14th element belongs to 110 - 120 class

The median value is then calculated as:

$\mathbf{Median = L + \frac{(n/2 -cf)}{f} \times h}$

Where:

$\mathbf{n/2 = 14}$

$\mathbf{L = 110}$ --- lower limit of the median class

$\mathbf{cf =11}$ ---- cumulative frequency of the class before the cumulative frequency

$\mathbf{f = 5}$ ---- frequency of the median class

$\mathbf{h = 10}$ --- the class size

So, we have:

$\mathbf{Median = 110 + \frac{14 - 11}{5} *10}$

$\mathbf{Median = 110 + \frac{3}{5} *10}$

$\mathbf{Median = 110 + 0.6 *10}$

$\mathbf{Median = 110 + 6}$

$\mathbf{Median = 116}$

Hence, the median is 116

(b) The lower quartile

Calculate the median class as follows

$\mathbf{Q_1 = \frac{1}{4}(n)}$

This gives

$\mathbf{Q_1 = \frac{1}{4}(28)th}$

$\mathbf{Q_1 =7th}$

The 7th element belongs to 100 - 110 class

The median value is then calculated as:

$\mathbf{Q_1 = L + \frac{(n/4 -cf)}{f} \times h}$

Where:

$\mathbf{n/4 = 7}$

$\mathbf{L = 100}$ --- lower limit of the lower quartile class

$\mathbf{cf =6}$ ---- cumulative frequency of the class before the cumulative frequency

$\mathbf{f = 5}$ ---- frequency of the median class

$\mathbf{h = 10}$ --- the class size

So, we have:

$\mathbf{Q_1= 100 + \frac{7 - 6}{5} *10}$

$\mathbf{Q_1= 100 + \frac{1}{5} *10}$

$\mathbf{Q_1= 100 + 0.2 *10}$

$\mathbf{Q_1= 100 + 2}$

$\mathbf{Q_1= 102}$

Hence, the lower quartile is 102

Read more about median and quartiles at:

https://brainly.com/question/3279114