We have given that X be the random variable normally distributed with mean =80 and standard deviation =5
X~N(80,52)
So, to find
P(65<X<73)
[Z~N(0,1)]
5.2.7
Let X be the random variable denoting mean =507 and standard deviation =122
To find
a) P(X<550)
=0.6378
approximately 63.78% sat score less than 550.
b)
here first find,
P(X>525)
=P(Z>0.147541)
=0.4414
n=1000 sample has taken, so 1000*0.4414=441.4 samples are approx greater than 525.
(442 samples in integer)
#########################part c
In this problem X~N(100,132)
a) P(X<66)
=P(Z<-2.615385)
=0.0045
b)
P(88<X<110)
=P(-0.9230769<Z<0.7692308)
=0.6011
c)
P(X>130)
=P(Z>2.307692)
=0.0105
Assume the random variable x is normally distributed with mean u = 80 and standard deviation...
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