Transcript Comparing Normal Distribution

Comparing normal distributions
To compare two sets of data we use z-scores. Using z-scores allows
us to compare the relative positions of two scores that are in different
distributions, that may have different means and standard deviations.
When we change our raw scores into standardised scores, or z-scores,
the problems of different means and standard deviations are
overcome. This is because each score is now expressed in relative
values.
The way we use the z-score will depend upon the question asked. If
the question wants the best result, we will use the biggest z-score.
If the question wants the worst result, we will use the smallest zscore.
If we want which score is closest to the average of it’s distribution,
we will use the z-score closest to zero.
Rearranging the z-score formula z  x  x to make
s
x the subject, we get:
x  sz  x
Example 1
Emma compares her results in French and Japanese. In French Emma
received 82%, the class average was 68% and the standard deviation
of the class was 8.45%. In Japanese Emma received 37 out of 50, the
class average was 28 and the standard deviation of the class was 5.3.
In which subject did Emma do best?
French
xx
z
s
82  68

8.45
Japanese
xx
z
s

 1.66 (2dp)
The subject Emma did best in was Japanese, as
her z-score was highest for Japanese.
37  28

5.3
 1.70 (2dp)
Example 2
Sam, a zoologist, is studying the weights of mice and elephants. The
breed of elephants she is studying has a mean weight of 20.6 tonnes
and a standard deviation of 3.2 tonnes. The breed of mouse Sam is
studying has a mean weight of 35.6 grams and a standard deviation of
2.8 grams. Which animal is more underweight, Dumbo the elephant
weighing 17.1 tonnes, or Mickey the mouse weighing 31.4 grams?
Dumbo
z
xx
s
17.1  20.6

3.2
 1.09 (2dp) 
Mickey
z
xx
s
31.4  35.6

2.8
 1.5
Mickey the mouse is the more underweight, as his
z-score was lower than Dumbo’s.
Example 3
Sinead and Paul compete in synchronised swimming teams in different states. Sinead’s
team, the lovely legs, competes in Victoria, where they say the judging is the hardest
in the country. The lovely legs scored 7.3 last weekend in a competition that averaged
6.78 and had a standard deviation of 0.42. Paul’s team, the fabulous floaters, competes
in Queensland. The fabulous floaters scored 8.4 two weeks ago in a competition that
averaged 7.43 and had a standard deviation of 0.61.
(a) Calculate the score the lovely legs would have got if they competed in the
Queensland competition, assuming they performed to the same level.
(b) In which state are the judges more consistent with their scores?
Calculate the z-score for the lovely legs.
Calculate the lovely legs score in the
Queensland competition.
xx
Lovely legs
z
s
7.3  6.78

0.42

 1.24 (2dp)
x  sz  x
 0.61 1.24  7.43
 8.19 (2dp)
b) Victoria had a standard deviation of 0·42. Qld had a standard deviation of 0·61.
 Victoria had the most consistent judges.
Today’s work
Exercise 8A pg 239
#6, 7, 9, 10
Yesterday’s work
Exercise 10-02
Q1→5 & 10