Probability question

Repair job     

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

i. probability that none of them are left-handed,

0.92^20 = 0.18869

 ii. probability that at most 2 are left-handed,

zero one or two are left handed

=           0.92^20          +   20 * 0.92^19 * 0.08  +    20C2 * 0.92^18  *  0.08^2

=     0.188693329     +      0.328162311     +               0.271090605

= 0.81648

  iii. standard deviation for the number of left-handed students

Don't know, I would have to research this one.

http://www.dummies.com/how-to/content/how-to-find-the-mean-variance-and-standard-deviati.html

p=0.08    q=0.92   n=20

I think this stays the same as before.

\(\sigma=\sqrt{np(1-p)}\\ \sigma=\sqrt{20*0.08*0.92}\\ \sigma\approx 1.21326\)

(b) If 50 classes of 20 students are randomly selected, what is the probability that 10 classes have no left-handed students? (10 have all right ahanders and the other classes do no have all right handers.

I think

50C10 * (0.92^20)^10 * (1-0.92^20)^40 = 0.137 

(all final answers are approximation. )

Nov 20, 2015 12:22 PM

Probabilities are the wrong way around I think Melody.

Probabilty that the first is'nt left handed is 0.92, the second also 0.92 etc.

Same mistake with the second and third parts.

Yes, you are right - I'll have to wear my glasses next time LOL    Thanks  

ii. Answer is 0.78794 not 0.81648