Statistics P6 Help needed

[imagine]

can anyone plzz help me with this

mean= np= 30*0.8= 24
var= npq= 30.0.8*0.2= 4.8

P(X<25)
and normalise it: P(X<24.5)
now u can solve it

oh btw, reason for next answer would be "calculating probability for many days takes more effort and consumes for time if normalisation method isn't used".

 

[parthrocks]

hey help me out plzz....with proper stepslz

 

[parthrocks]

help khan!!

 

[parthrocks]

help khan!!View attachment 10538

I didnt get the second part......why is the probab 0.05

 

[imagine]

hey help me out plzz....with proper stepslz
View attachment 10534

(i) mean= (5/3) s.d.

P(Z<(2mew-mew)/sd)
by substituting the above relation given we get : fy(5/3)
and the final value is 0.952


(ii) in P(X<mew/3), just do the solving like we are supposed to do...( u understand, right?)
for 0.8524, from the normal distribution table, write fy(1.047)
solving both sides we get:
fy(-2mew/3sd)= fy(1.047)
or, -2 mew= 1.047*3 sd
and finally, mew= -1.57 sd ( answer)

hope this will help u

 

[imagine]

I didnt get the second part......why is the probab 0.05

can I help u??

 

[shanky631]

can I help u??

hey dude when do we take negative value of z. i don't understand.

 

[parthrocks]

can I help u??

Ya sure....sorry for mentioning the name up.......bt u are cool man.....thank u

 

[parthrocks]

heybro and can u plzzz tell me the steps in detail for this....i know its easy bt a silly mistake here maybe.....IMAGINE......PLZZ

 

[imagine]

hey dude when do we take negative value of z. i don't understand.

in condition when the given probability is not available in the table...I guess you didn't understand this
so as an example, we are given a probability = 0.10

0.10 is nt available in distribution table so do= 1-0.1= 0.9
now the z value for 0.9 can be obtained which is 1.282 and thus in that obtained value of z we put minus(-) sign b4 the value of z i.e. -1.282

hope u understand well

 

[shanky631]

in condition when the given probability is not available in the table...I guess you didn't understand this
so as an example, we are given a probability = 0.10

0.10 is nt available in distribution table so do= 1-0.1= 0.9
now the z value for 0.9 can be obtained which is 1.282 and thus in that obtained value of z we put minus(-) sign b4 the value of z i.e. -1.282

hope u understand well

finally. hats off to u. Now i get it. You are great.

 

[parthrocks]

in condition when the given probability is not available in the table...I guess you didn't understand this
so as an example, we are given a probability = 0.10

0.10 is nt available in distribution table so do= 1-0.1= 0.9
now the z value for 0.9 can be obtained which is 1.282 and thus in that obtained value of z we put minus(-) sign b4 the value of z i.e. -1.282

hope u understand well

IS IT ALWAYS THE CASE

 

[parthrocks]

WHY THERE IS NO CONTINUITY CORRECTION???

 

[math]

continuity correction is wen suitable approximation is mention in the question

 

[parthrocks]

continuity correction is wen suitable approximation is mention in the question

yA....ACTUALLY...IT DOES WORK....and hey in have posted sum doubts...can u solve it if possible

 

[shanky631]

WHY THERE IS NO CONTINUITY CORRECTION???

which paper??

 

[parthrocks]

which paper??

Its in almost all cases O/N 2010/63

 

[shanky631]

Its in almost all cases O/N 2010/63

which question.

 

[imagine]

IS IT ALWAYS THE CASE

YEAH!

 

[imagine]

finally. hats off to u. Now i get it. You are great.

anytime