REVIEW Normal Distribution. Characterizing a Normal Distribution. To completely characterize a normal distribution, we need to know only 2 things: The mean --- The standard deviation --- . HAND CALCULATIONS FOR THE NORMAL DISTRIBUTION.
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REVIEW
Normal Distribution
To completely characterize a normal distribution, we need to know only 2 things:
Some tables give probability of falling between 0 and a positive z value
a X
Normal Curve with X and Z ScalesProbabilities from 0 to z
µ a X
0 z Z
A cumulative normal table gives the probability of falling between -∞ and any z value
a X
Curve with X and Z ScalesCumulative Probabilities from -∞ to z
µ a X
0 z Z
P(X<a) or P(Z<z) = area between -∞ and a (or z)
EXCEL:
=NORMDIST(a,µ,σ,TRUE) or
=NORMSDIST(z)
P(X>a) or P(Z>z) = area between a (or z) and +∞
EXCEL:
=1-NORMDIST(a,µ,σ,TRUE) or
=1-NORMSDIST(z)
P(a<X<b) or P(za<Z<zb)
EXCEL:
=NORMDIST(b,µ,σ,TRUE) - NORMDIST(a,µ,σ,TRUE)or =NORMSDIST(zb) - NORMSDIST(za)
Determining the x value such that the probability of getting a value less than x is p
EXCEL:
= NORMINV(p,µ,σ) or
= µ + NORMSINV(p)*σ
FROM TABLE
.7734
.4332
335
σ = 20
320 X
0.75
0 Z
EXCEL
=NORMDIST(335,320,20,TRUE)
OR =NORMSDIST(.75)
FROM TABLE
.5987
1 - .5987 =
.4013
.4332
325
σ = 20
320 X
0.25
0 Z
EXCEL
=1-NORMDIST(325,320,20,TRUE)
OR =1-NORMSDIST(.25)
.9192 - .1977 =
.7215
FROM TABLE
.9192
FROM TABLE
.1977
.4332
303 348
σ = 20
320 X
-0.85
1.40
0 Z
EXCEL
=NORMDIST(348,320,20,TRUE)-NORMDIST(303,320,20,TRUE)
OR =NORMSDIST(1.40)-NORMSDIST(-0.85)
.7500 is to the
left of x
x = 320 + .67(20)
.4332
The closest value is .7486 which
corresponds to a z-value of 0.67.
x
333.4
0.67
σ = 20
Try to find .7500 in the middle
of the cumulative normal table.
320 X
0 Z
EXCEL
=NORMINV(.75,320,20)
OR =320 + NORMSINV(.75)*20
Thus,
1-.8500 = .1500
is to the left of x
.8500 is to the
right of x
x = 320 + (-1.04)(20)
.4332
The closest value is .1492
which corresponds to a
z-value of -1.04
299.2
x
-1.04
σ = 20
Try to find .1500 in the middle
of the cumulative normal table.
320 X
0 Z
EXCEL
=NORMINV(.15,320,20)
OR =320 + NORMSINV(.15)*20