Diffusion is the movement of particles from a high to lower concentration. Osmosis is the diffusion of water across a membrane. Active transport moves particles from low to higher concentration.

The investigation cannot be based on changes in mass from just one potato cylinder. In scientific tests, we must ensure a number of examples are used to allow for anomalous results and variation.

Percentage changes in mass must be calculated for each cylinder and mean calculated.

A graph is plotted of change in mass, in per cent, against concentration of sucrose. Concentrations of solutions are shown as percentages. A larger number means a higher concentration.

Where potato cylinders have gained in mass, the change will be positive. Where potato cylinders have decreased in mass, the change will be negative.

Concentration of sucrose % | Average change in mass (%) |
---|---|

0 | +26.8 |

20 | +5.0 |

40 | −7.7 |

60 | −17.9 |

80 | −26.0 |

100 | −31.4 |

Where the plotted line crosses the horizontal axis at 0 per cent change in mass, the sucrose concentration of the solution is equal to the concentration of the contents of the potato cells.

This can be identified on the graph as the point which shows no net movement of water by osmosis, which would be represented by a change in mass.

- Question
What is the concentration of solutes in the cells of the potato in this investigation?

29%

This value is where the line crosses the x-axis.

The concentration of dissolved solutes in the cells of different potatoes will vary slightly from potato to potato. If we have a set of data for a range in concentrations, we can look at the range, and the mean - but these do not tell us whether data are evenly spread or whether they are clustered together within a certain range.

Scientists use percentiles to divide a set of data into 100, and look to see where the data lie within these divisions.

The median is the point in a set of data where 50 per cent of the data fall above this value, and 50 per cent below it. This is the 50^{th} percentile.

The 75^{th} percentile is where 75 percent of the data fall below this value.

There are several methods of finding a percentile. The simplest is the nearest rank method. As with finding the median of a set of data, begin by putting the data into order.

For a range of values for the concentration of potato cell sap:

0.27 | 0.32 | 0.25 | 0.24 | 0.28 | 0.31 | 0.30 | 0.26 | 0.29 | 0.29 | 0.31 | 0.35 | 0.21 | 0.28 | 0.28 | 0.26 |

0.35 | 0.22 | 0.27 | 0.26 | 0.24 | 0.23 | 0.39 | 0.28 | 0.29 | 0.27 | 0.26 | 0.25 | 0.30 | 0.27 | 0.25 | 0.26 |

Arranged in order:

0.21 | 0.22 | 0.23 | 0.24 | 0.24 | 0.25 | 0.25 | 0.25 | 0.26 | 0.26 | 0.26 | 0.26 | 0.26 | 0.27 | 0.27 | 0.27 |

0.27 | 0.28 | 0.28 | 0.28 | 0.28 | 0.29 | 0.29 | 0.29 | 0.30 | 0.30 | 0.31 | 0.31 | 0.32 | 0.35 | 0.35 | 0.39 |

To find, for example the 50^{th} percentile, first find the rank:

So the 50^{th} percentile will be the 16^{th} number in the ordered data set, starting from the left. The 50^{th} percentile is 0.27.

0.21 | 0.22 | 0.23 | 0.24 | 0.24 | 0.25 | 0.25 | 0.25 | 0.26 | 0.26 | 0.26 | 0.26 | 0.26 | 0.27 | 0.27 | 0.27 |

0.27 | 0.28 | 0.28 | 0.28 | 0.28 | 0.29 | 0.29 | 0.29 | 0.30 | 0.30 | 0.31 | 0.31 | 0.32 | 0.35 | 0.35 | 0.39 |

In instances where the ordered rank is not a whole number, you should round the number up.

- Question
Find the 90

^{th}percentile for the same set of data.0.32 mol dm

^{−3}.This method will only give percentiles as numbers that exist in the data set.

In other methods, percentiles can be interpolated for values that don't exist in the data set.