Download Normal / Z-Score Powerpoint

standard deviations
standard deviation
spread
distance
difference
Sample Data
Consider the following test scores for a small class:
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
73
Jenny’s score is noted in red. How did she perform on this
test relative to her peers?
6| 7
7 | 2334
7 | 5777899
8 | 00123334
8 | 569
9 | 03
6| 7
7 | 2334
7 | 5777899
8 | 00123334
8 | 569
9 | 03
Her score is “above average”...
but how far above average is it?
a standardized value
a z-score
no
comparable
change the SHAPE
the center….your new mean is ZERO
the spread….your new standard deviation is ONE
N (0, 1)
Calculating z-scores
Consider the test data and Jenny’s score.
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
According to Minitab, the mean test score was 80 while the standard
deviation was 6.07 points.
Jenny’s score was above average. Her standardized z-score is:
x  80 86  80
z

 0.99
6.07
6.07
Jenny’s score was almost one full standard deviation above the mean.
What about Kevin: x=
73
Calculating z-scores
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
6| 7
7 | 2334
7 | 5777899
8 | 00123334
8 | 569
9 | 03
Jenny: z=(86-80)/6.07
z= 0.99 {above average = +z}
Kevin: z=(72-80)/6.07
z= -1.32 {below average = -z}
Katie: z=(80-80)/6.07
z= 0
{average z = 0}
73
Comparing Scores
Standardized values can be used to
compare scores from two different
distributions.
Statistics Test: mean = 80, std dev = 6.07
Chemistry Test: mean = 76, std dev = 4
Jenny got an 86 in Statistics and 82 in
Chemistry.
On which test did she perform better?
Statistics
86  80
z
 0.99
6.07
Chemistry
82  76
z
 1.5
4
Although she had a lower score, she performed relatively better in
Chemistry.


-∞ = -EE99
∞=EE99
2:Normalcdf
(lower bound,
upper bound)
N(500,100)

What percentile will you be in if you scored
a 720 on the SAT??
N(500,100)

What percent of students scored between
550 and 630 on the SAT??
under the normal curve
ONE
SYMMETRIC!!!
One of the side effects of flooding a lake in northern boreal forest
areas (e.g. for a hydro-electric project) is that mercury is leached
from the soil, enters the food chain, and eventually contaminates
the fish. The concentration in fish will vary among individual fish
because of differences in eating patterns, movements around the
lake, etc. Suppose that the concentrations of mercury in individual
fish follows an approximate normal distribution with a mean of
0.25 ppm and a standard deviation of 0.08 ppm. Fish are safe to
eat if the mercury level is below 0.30 ppm. What proportion of
fish are safe to eat?
Marks on a Chemistry test follow a normal distribution
with a mean of 65 and a standard deviation of 12.
Approximately what percentage of the students have
scores below 50?
inverse
% or percentile
Next Question…

What is your test score above if you are in
the top 20%?
2.1 Summary
We can describe the overall pattern of a
distribution using a density curve.
The area under any density curve = 1.
This represents 100% of observations.
Areas on a density curve represent % of
observations over certain regions.
An individual observation’s relative
standing can be described using a zscore or percentile rank.
x  mean
z
standard deviation
END OF NOTES
Homework:
Packet p. 7 #1-8, omit 4
Make sure you understand #2.2!!
Warm Up – Day 8
1)
2)
3)
4)
What is the z-score formula?
Can you say and spell the Greek symbols
in the z score formula?
The average number of parents that
attend Meet the Teacher night at GHHS is
323 with a standard deviation of 5.6.
Interpret the standard deviation in context.
(*This stat was created by your teacher!)
What is wrong with this statement: I have
a z score of -4.3 points.
Work on Matching
Activity
Answers on BB
END OF NOTES
•
•
Start HOMEWORK:
• Packet page 6 #1 - 8
Remember packet 6 is due
Thursday!