14 thoughts on “BscCSSemester3 Probability distribution and sampling theory Normal distribution PART4”
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Affiliated to the University of Calicut
Affiliated to the University of Calicut
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Ans 4) p(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
Ans 4) p(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
And 4) p(-1<z<2)
=P(-1<z<0)+p(0<z<2)
=0.3413+0.4772
=0.8185
And 4) p(-1<z<2)
=P(-1<z<0)+p(0<z<2)
=0.3413+0.4772
=0.8185
And 4)p(-1<z<2)
=p(-1<z<0)+p(0<z<2)
=0.3413+0.4772
=0.8185
P(38<x<50)
P(-1<z<2)
P(-1<z<0) +P(0<z<2)
=0.3413+0.4772
=0.8185
Ans 4) p(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
Ans 4) p(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
Ans 4) p(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
P(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
Ans 4) p(-1<z<2)
=P(-1<z<0)+P(0<z<2)
=0.3413+0.4772
=0.8185
Ans 4) p(-1<z<2)
=P(-1<z<0)+p(0<z<2)
=0.3413+0.4772
=0.8185
P(-1<z<2)
=P (-1<z<0) +p(0<z<2)
=0.3413+0.4772
= 0.8185
P(-1<z<2)
=P (-1<z<0) +p(0<z<2)
=0.3413+0.4772
= 0.8185