Question

It has been estimated that only about 40% (p = 0.4) of California residents have adequate earthquake supplies. Suppose you randomly survey 50 California residents. We are interested in the number who have adequate earthquake supplies. What is the probability that exactly 30 residents will have the needed supplies?

Answer 1

Solution:

p = 0.4

1 - p = 0.6

n = 50

X follows Binomial(50, 0.4)

then,

n*p = 50 * 0.4 = 20

n(1- p) = 50 * 0.6 = 30

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

According to normal approximation binomial,

X Normal

Mean = = n*p = 20

Standard deviation = =n*p*(1-p) = [50*0.4*06] = 3.46410161514

We using continuity correction factor

P(X = a) = P( a - 0.5 < X < a + 0.5)

So

P(Exactly 30)

= P(X = 30)

= P(29.5 < X < 30.5)

= P(X < 30.5) - P(X < 29.5)

=  P[(X - )/ <  (30.5 - 20)/3.46410161514] -   P[(X - )/ <  (29.5 - 20)/3.46410161514]

= P[Z < 3.03] - P[Z < 2.74]

= 0.9988 - 0.9969 ..Use z table

= 0.0019