Testee A: 1 1 1 1 1 = 5 points

Testee B: 1 0 1 1 1 = 4 points

Testee C: 1 0 0 1 1 = 3 points

Testee D: 0 1 1 1 0 = 3 points

Testee E: 1 0 1 0 0 = 2 points

So, item 1 is solved by 4 testees: A, B, C, E. This means that S=4 according
to **http://ivec.ultimaiq.net/quality.htm**

Among top 4(S) testees item 1 is solved by A, B and C, so R=3.

The quality of item 1 is so R/S=3/4=75%.

Item 5 is problematic here. It is solved by A, B and C, which means S=3.

But top 3 testees do not exist because testees C and D share 3 points each.

So, I will take S=4 here, and then again R=3, and again 3/4=75%.

Item 2: 1/2=50%

Item 3: 3/4=75%

Item 4: 4/4=100%

Try to clarify this to yourself! If S is not an integer, round to the nearest integer.

So we have 4 good items (1,3,4,5) and 1 average (2).

Solvability of item is just an average score. For item 1 it is 4/5=0.8,

and for all items we have 0.8, 0.4, 0.6, 0.8, 0.6 in order.

Sorting by solvability we have the following table (I call it table A below):

item |
solvability |
quality |

4 |
0.8 |
100% |

1 |
0.8 |
75% |

3 |
0.6 |
75% |

5 |
0.6 |
75% |

2 |
0.4 |
50% |

Now, let us assume that we have the following norm:

raw score |
IQ |

1 |
130 |

2 |
140 |

3 |
150 |

4 |
160 |

5 |
170 |

IQ 130-150 range is raw score 1-3 range, so for quality in that range we take
only the first three rows from table A:

(100%+75%+75%)/3=83.3%.

140-160 range is raw 2-4, so we take rows 2-4 from table A:

(75%+75%+75%)/3=75%.

Quality in the 150-170 range is thus (75%+75%+50%)/3=66.7%.