Question

Suppose that 10% of students are left-handed.

a) In a study group of 6 students, find the exact probability that at most 1 of them is left-handed.

b) In a classroom of 60 students, find an approximate probability that at most 10 of them are left-handed.

Answer 1

a)

Using binomial distribution,

P(X) = ⁿC_x p^x (1-p)^n-x

P( X <= 1) = P( X = 0) + P( X = 1)

= ⁶C₀ 0.10⁰ 0.90⁶ + ⁶C₁ 0.10¹ 0.90⁵

= 0.8857

b)

Mean = np = 60 * 0.10 = 6

Standard deviation = sqrt(np(1-p))

= sqrt(60 * 0.10 * 0.90)

= 2.3238

Using normal approximation,

P( X <= x) = P( Z < x+0.5 - mean / SD)

So,

P( X < 10) = P( Z < 10.5 - 6 / 2.3238)

= P( Z < 1.9365)

0.9736