It is noted that 8% of students are left handed. If 20 (TWENTY) students are randomly selected, calculate the

Hi Why

I am on my phone do layout is difficult.

 

P of no lefts = 0.92^20

 

P ( at most 2 lefts )

=P ( no  lefts )+P ( 1 lefts )+P ( 2 lefts )

=20C0×0.92^20+20C1×0.08×0.92^19+20C2×0.08^2×0.92^18

Etc

 

Mean=20×0.08

SD=sqrt( 20×0.08×0.92  )

thank u for your helping!!!

i got it!

 

and then (b)question is what mean ,i dont understand and i dont know how to do it

Sorry, I am not sure how to do part b - I am still considering . ://

I may answer it later.

Ok I think I have it now :))

 

It is noted that 8% of students are left handed. If 20 (TWENTY) students are randomly selected,

 

P( no one in a class of 20 is left handed) = 0.92^20            =  0.18869

P( one or more in a class of 20 is left handed = 1-0.18869 =   0.81131

 

(b) If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN) classes have no left-handed students?

 

P(10 classes of 20 students have no left-handed students)   

= 50C10 * 0.18869^10 * 0.81131^40

 

nCr(50,10) * 0.18869^10 * 0.81131^40 = 0.1369710474047608673359021944148508674768838

 

The probability is 0.137