Question

66% of all Americans are homeowners. If 40 Americans are randomly selected, find the probability that

a. Exactly 27 of them are homeowners.
b. At most 25 of them are homeowners.
c. At least 24 of them are homeowners.
d. Between 24 and 29 (including 24 and 29) of them are homeowners.

Answer 1

Mean = n * P = ( 40 * 0.66 ) = 26.4
Variance = n * P * Q = ( 40 * 0.66 * 0.34 ) = 8.976
Standard deviation = = 2.996

Part a)

P ( X = 27 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 27 - 0.5 < X < 27 + 0.5 ) = P ( 26.5 < X < 27.5 )

P ( 26.5 < X < 27.5 )
Standardizing the value

Z = ( 26.5 - 26.4 ) / 2.996
Z = 0.03
Z = ( 27.5 - 26.4 ) / 2.996
Z = 0.37
P ( 0.03 < Z < 0.37 )
P ( 26.5 < X < 27.5 ) = P ( Z < 0.37 ) - P ( Z < 0.03 )
P ( 26.5 < X < 27.5 ) = 0.6432 - 0.5133
P ( 26.5 < X < 27.5 ) = 0.1299

Part b)

P ( X <= 25 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 25 + 0.5 ) = P ( X < 25.5 )

P ( X < 25.5 )
Standardizing the value

Z = ( 25.5 - 26.4 ) / 2.996
Z = -0.3

P ( X < 25.5 ) = P ( Z < -0.3 )
P ( X < 25.5 ) = 0.3821

Part c)

P ( X >= 24 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 24 - 0.5 ) =P ( X > 23.5 )

P ( X > 23.5 ) = 1 - P ( X < 23.5 )
Standardizing the value

Z = ( 23.5 - 26.4 ) / 2.996
Z = -0.97

P ( Z > -0.97 )
P ( X > 23.5 ) = 1 - P ( Z < -0.97 )
P ( X > 23.5 ) = 1 - 0.166
P ( X > 23.5 ) = 0.834

Part d)

P ( 24 <= X <= 29 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 24 - 0.5 < X < 29 + 0.5 ) = P ( 23.5 < X < 29.5 )

P ( 23.5 < X < 29.5 )
Standardizing the value

Z = ( 23.5 - 26.4 ) / 2.996
Z = -0.97
Z = ( 29.5 - 26.4 ) / 2.996
Z = 1.03
P ( -0.97 < Z < 1.03 )
P ( 23.5 < X < 29.5 ) = P ( Z < 1.03 ) - P ( Z < -0.97 )
P ( 23.5 < X < 29.5 ) = 0.8496 - 0.1665
P ( 23.5 < X < 29.5 ) = 0.6831