z distribution

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Z-score
(Standardized Normal Deviate)
The Standard Normal Distribution
The Standard Normal distribution follows a normal
distribution and has mean 0 and standard deviation 1
Notice that the distribution is perfectly symmetric about 0.
If a distribution is normal but not standard, we can convert a
value to the Standard normal distribution table by first by

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finding how many standard deviations away the number is
from the mean.
The z-score
[Q: Write short notes on: Z-score. (BSMMU, MD Radiology,
July, 2010)]
The number of standard deviations from the mean is called
the z-score and can be found by the formula
x
z




Explanation:
How would you compare the number of books in ‘library A’
to the number of books in other libraries? What type of
statistic would you use to actually give information about
‘library A’, short of counting every book in several different
libraries?
You would use a standard score. Standard scores are
calculated in order to describe the extent of variation of a
value as it would be compared to another value. They are the
most convenient way to compare similar or different values
by a similar scale.
If ‘library A’ contains 250 books, you won’t be able to tell how
that number compares to other libraries, such as that of the
average college professor. To more effectively describe
‘library A’ in comparison to other libraries, we calculate a
standard score or z-score. The standard score number

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describes the location of a particular case in a distribution:
whether it is above or below the average and by how much.
The z-score is expressed in standard deviation units, so it
always gives you an idea of the magnitude of the difference
compared to the distribution of values for the whole
population.
To calculate the z-score of a value x from a distribution one
needs the mean  and the standard deviation  for that
distribution. The formula reads:




x
z
In the library example, if book ownership among college
professors has a mean  of 150 and a standard deviation  of
50, the z-score for library A (consisting of 250 books) is:
2
50
150
250
z 


The z-score for library A is positive, indicating it has more
books than average. It tells us that library A is two standard
deviations over the mean. Now you probably will remember
the empirical rule for bell-shaped distribution, which tells us
that
 approximately 68% of all observations is less than 1
standard deviation away from the mean, and
 approximately 95% of all observations is less than 2
standard deviations away from the mean,
so we now have a pretty good idea about the magnitude of
library A compared to other ones. This particular library is

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quite extensive, compared to the ones of other college
professors.
The pictures below show us this empirical rule.

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To conclude: z-scores inform us in a standardized way about
the position of an observation within a certain distribution.
Example
Find the z-score corresponding to a raw score of 132 from a
normal distribution with mean 100 and standard deviation 15.
Solution
We compute
132 - μ
z = = 2.133
15
Example
A z-score of 1.7 was found from an observation coming from
a normal distribution with mean 14 and standard deviation 3.
Find the raw score.
Solution
We have
x - μ
1.7 =
3
To solve this we just multiply both sides by the denominator
3,

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(1.7)(3) = x - 14
5.1 = x - 14
x = 19.1
The z-score and Area
Often we want to find the probability that a z-score will be
less than a given value, greater than a given value, or in
between two values. To accomplish this, we use the table
from the textbook and a few properties about the normal
distribution.
Example
Find
P(z < 2.37)

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Solution
We use the table. Notice the picture on the table has shaded
region corresponding to the area to the left (below) a z-
score. This is exactly what we want. Below are a few lines of
the table.
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890
2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916
2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936
The columns corresponds to the ones and tenths digits of the
z-score and the rows correspond to the hundredths digits.
For our problem we want the row 2.3 (from 2.37) and the row
.07 (from 2.37). The number in the table that matches this is
.9911.
Hence
P(z < 2.37) = .9911
Example
Find
P(z > 1.82)
Solution

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In this case, we want the area to the right of 1.82. This is not
what is given in the table. We can use the identity
P(z > 1.82) = 1 - P(z < 1.82)
reading the table gives
P(z < 1.82) = .9656
Our answer is
P(z > 1.82) = 1 - .9656 = .0344
Example
Find P(-1.18 < z < 2.1)
Solution

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Once again, the table does not exactly handle this type of
area. However, the area between -1.18 and 2.1 is equal to the
area to the left of 2.1 minus the area to the left of -1.18. That
is
P(-1.18 < z < 2.1) = P(z < 2.1) - P(z < -1.18)
To find P(z < 2.1) we rewrite it as P(z < 2.10) and use the table
to get
P(z < 2.10) = .9821.
The table also tells us that
P(z < -1.18) = .1190
Now subtract to get
P(-1.18 < z < 2.1) = .9821 - .1190 = .8631

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