Solution :
Let x be student will be left handed
P = 0.09
Using the normal approximation to binomial distribution,
a). n = 108,
μ = np = 9.72
[tex]$\sigma = \sqrt{np(1-p)}$[/tex]
[tex]$=\sqrt{8.8452}$[/tex]
= 2.9741
Required probability,
P(x=8) = P(7.5 < x < 8.5)
[tex]$=P\left(\frac{7.5-9.72}{2.9741}< \frac{x-\mu}{\sigma}< \frac{8.5-9.72}{2.9741}\right)$[/tex]
[tex]$=P(-0.75 < z < -0.41)$[/tex]
Using z table,
= P(z<-0.41)-P(z<-0.75)
= 0.3409-0.2266
= 0.1143
b). P(x=12) = P(11.5 < x < 12.5)
[tex]$=P\left(\frac{11.5-9.72}{2.9741}< \frac{x-\mu}{\sigma}< \frac{12.5-9.72}{2.9741}\right)$[/tex]
[tex]$=P(0.60 < z < 0.94)$[/tex]
Using z table,
= P(z< 0.94)-P(z< 0.60)
= 0.8294 - 0.7257
= 0.1006
