| Tag | Value |
|---|---|
| file | Descriptive-statistics_vufsw-zscore-1084-en_vufsw-zscore-1084-en |
| name | vufsw-zscore-1084-en |
| section | descriptive statistics/score interpretation/z-score |
| type | schoice |
| solution | TRUE, FALSE, FALSE, FALSE, FALSE |
| Type | calculation |
| Program | calculator |
| Language | English |
| Level | statistical literacy |
The next question concerns the Glastonbury Festival in England, where
the physical hygiene of visitors was measured on a scale of 0 to 4.
See Table 1 below.
Table 1: Descriptive statistics of physical hygiene of visitors Glastonbury Festival.
|
|
Hygiene (Day 1 of Glastonbury
Festival) |
Hygiene (Day 2 of Glastonbury
Festival) |
Hygiene (Day 3 of Glastonbury Festival) |
|
N |
810 |
810 |
810 |
|
Missing |
0 |
0 |
0 |
|
Mean |
1.77 |
0.96 |
0.98 |
|
Median |
1.79 |
0.79 |
0.76 |
|
Std. Deviation |
0.69 |
0.72 |
0.71 |
|
Percentile 25 |
1.31 |
0.41 |
0.44 |
|
Percentile 50 |
1.79 |
0.79 |
0.76 |
|
Percentile 75 |
2.23 |
1.35 |
1.55 |
Suppose one of the visitors has a physical hygiene score of 0.88 on day three. What is his / her standardized score (z-score)?
The formula is: z = (observation - mean) / standard deviation
Thus, z = (0.88 - 0.98) / 0.71 = -0.14
Language
English
Levels of Difficulty
Medium
M&T BIS
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M&T Descriptive statistics
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M&T Distributions
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