## Related questions with answers

The following table gives the top ten countries in the world whose populations have the highest concentration of type $\mathrm{O}^{+}$blood:

Country | Saudi Arabia | Iceland | Ireland | Taiwan | Australia |
---|---|---|---|---|---|

Population (%) | $48.0$ | $47.6$ | $47.0$ | $43.9$ | $40.0$ |

Country | Hong Kong | Italy | Netherlands | Canada | South Africa |
---|---|---|---|---|---|

Population (%) | $40.0$ | $39.5$ | $39.0$ | $39.0$ | $38.0$ |

Find the mean (average), median, and mode of these concentrations of type $\mathrm{O}^{+}$blood in these ten countries.

Solution

VerifiedThe mean is given by:

$\begin{equation} \begin{aligned} \overline{x}& = \dfrac{x_1 + x_2 + ... + x_n}{n}\\ &= \quad \dfrac{48.0+ 47.6+ 47.0 + 43.9+ 40.0+ 40.0+ 39.5 + 39.0 + 39.0 + 38.0}{10} \\ & = \dfrac{422}{10}\\ & = 42.2 \end{aligned} \end{equation}$

The mean of these concentrations of type $\text{O}^+$ blood in these ten countries is $42.2 \%$.

To find the median, we arrange the values in increasing order.

$38.0,\quad 39.0,\quad 39.0 ,\quad 39.5,\quad 40,\quad 40,\quad 43.9,\quad 47,\quad 47.6,\quad 48$

We have $10$ terms, the median is the average of the $5th$ and $6th$ terms.:

$\begin{equation} \begin{aligned} \text {Median}& = \dfrac{40+ 40}{2}\\ &= \dfrac{80}{2}\\ &= 40 \end{aligned} \end{equation}$

The median of these concentrations of type $\text{O}^+$ blood in these ten countries is $40 \%$.

Mode: $39$ and $40$ (because each number occurs two times)

The modes of these concentrations of type $\text{O}^+$ blood in these ten countries are $39\%$ and $40\%$ .

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