## 14 thoughts on “BscCSSemester3 Probability distribution and sampling theory Normal distribution PART4”

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# BscCSSemester3 Probability distribution and sampling theory Normal distribution PART4

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14 thoughts on “BscCSSemester3 Probability distribution and sampling theory Normal distribution PART4”

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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