Question

60% of all Americans are home owners. If 31 Americans are randomly selected, find the probability that

a. Exactly 18 of them are are home owners.

b. At most 18 of them are are home owners.

c. At least 19 of them are home owners.

d. Between 15 and 21 (including 15 and 21) of them are home owners.

Answer 1

Solution:-

Given that ,

random sample size=n = 31, p = 0.60

q = 1 - p = 1 - 0. = 0.4

Let x be a random variable of binomial distribution with n,p

For binomial distribution , P(x = r) = nCr*p^r*q^(n-r)

a. Exactly 18 of them are are home owners:

P(x = 18) = 31C18*0.6^18*0.4^13 = 0.1406

b. At most 18 of them are are home owners.

Since n = 31 > 30, we use Normal approximation

Mean = n * P = ( 31 * 0.6 ) = 18.6
Variance = n * P * Q = ( 31* 0.6 * 0.4 ) = 7.4

Standard deviation = sqrt ( variance ) = 2.7276

Using continuity correction
P ( X < n + 0.5 ) = P ( X < 18 + 0.5 ) = P ( X < 18.5 )

Standardizing the value

Z = ( 18.5 - 18.6 ) / 2.7276
Z = -0.04

P ( X < 18.5 ) = P ( Z < -0.04 )
P ( X < 18.5 ) = 0.4840

c. At least 19 of them are home owners.

P ( X >= 19 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 19 - 0.5 ) =P ( X > 18.5 )

Standardizing the value

Z = ( 18.5 - 18.6 ) / 2.7276
Z = -0.04

P ( X > 18.5 ) =1-P ( X < 18.5 ) = 1-P ( Z < -0.04 )
                        =1-P ( X < 18.5 ) =1- 0.4840

P ( X > 18.5 )=0.516

d. Between 15 and 21 (including 15 and 21) of them are home owners.

P ( 15 <= X <= 21 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 15 - 0.5 < X < 21+ 0.5 ) = P ( 14.5 < X < 21.5 )

Standardizing the value

Z = ( 14.5 - 18.6) / 2.7276
Z = -1.50
Z = ( 21.5 - 18.6 ) / 2.7276
Z =1.06

P ( - 1.50 < Z < 1.06 )
P ( 14.5 < X < 21.5 ) = P ( Z < 1.06 ) - P ( Z < -1.50 )
P ( 14.5 < X < 21.5 ) = 0.8554 - 0.0668
P ( 14.5 < X < 21.5 ) = 0.7886