Find the probability that the Normal random variable with mean 20 and standard deviation 3.2 will...

Find the probability that the Normal random variable with mean 20 and standard deviation 3.2 will generate an outlier (outside the inner fences) observation. Remember that the lower (upper) inner fence is 1.5*IQR below (above) the first (third) quartile.

a.	0.0035
	b.	0.0051
	c.	0.0058
	d.	0.0062
	e.	0.0070

Expert Solution

µ=   20
σ =    3.2
P(X≤x) =   0.2500

Z value at    0.25   =   -0.6745   (excel formula =NORMSINV(   0.25   ) )
z=(x-µ)/σ
so, X=zσ+µ=   -0.674   *   3.2   +   20
X   =   17.84 (answer)

so, first quartile, Q1 = 17.84

--------------------------

P(X≤x) =   0.7500

Z value at    0.75   =   0.6745   (excel formula =NORMSINV(   0.75   ) )
z=(x-µ)/σ
so, X=zσ+µ=   0.674   *   3.2   +   20
X   =   22.16 (answer)

so, third quartile, Q3 = 22.16

--------------

IQR = Q3-Q1 = 22.16-17.84 = 4.32

range of inner fence = (Q1 - 1.5IQR , Q3+1.5IQR) = 11.36 , 28.64

so,

P (   11.3600   < X <   28.6400   )
=P( (11.36-20)/3.2 < (X-µ)/σ < (28.64-20)/3.2 )

P (    -2.700   < Z <    2.700   )
= P ( Z <    2.700   ) - P ( Z <   -2.700   ) =    0.9965   -    0.0035   =    0.9930

P(outside the inner fence) = 1 - 0.9930 = 0.0070

please revert for doubt..

orchestra answered 2 years ago

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