Question

If 10% of the general population smokes, what is the approximate probability that fewer than 100 in a random sample of 928 people are smoker?

Answer 1

Solution:

Given that,

p = 10% = 0.10

1 - p = 1 - 0.10 = 0.90

n = 928

Here, BIN (928 , 0.10)

then,

n*p = 928 * 0.10 = 92.8

n(1- p) = 928 * 0.90 = 835.2

Since both are greater than 5 , we can use normal approximation to binomial.

According to normal approximation binomial,

X Normal

Mean = = n*p =  92.8

Standard deviation = =n*p*(1-p) = [928*0.10*0.90]  = 9.13892772704

Now ,

P( fewer than 100 )

= P(X < 100)

= P[(X - )/ <  (100 - )/]

= P[Z <  (100 - 92.8)/9.13892772704]

= P[Z < 0.79]

= 0.7852 ... ( use z table)

Required probability =  0.7852