Question

63% of all Americans are home owners. If 34 Americans are randomly selected, find the probability that

a. Exactly 23 of them are are home owners.

b. At most 21 of them are are home owners.

c. At least 23 of them are home owners.

d. Between 21 and 26 (including 21 and 26) of them are home owners.

Answer 1

Mean = n * P = ( 34 * 0.63 ) = 21.42
Variance = n * P * Q = ( 34 * 0.63 * 0.37 ) = 7.9254
Standard deviation =   = 2.8152

Part a)

P ( X = 23 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 23 - 0.5 < X < 23 + 0.5 ) = P ( 22.5 < X < 23.5 )

P ( 22.5 < X < 23.5 )
Standardizing the value

Z = ( 22.5 - 21.42 ) / 2.8152
Z = 0.38
Z = ( 23.5 - 21.42 ) / 2.8152
Z = 0.74
P ( 0.38 < Z < 0.74 )
P ( 22.5 < X < 23.5 ) = P ( Z < 0.74 ) - P ( Z < 0.38 )
P ( 22.5 < X < 23.5 ) = 0.77 - 0.6494
P ( 22.5 < X < 23.5 ) = 0.1206

Part b)

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

P ( X < 21.5 )
Standardizing the value

Z = ( 21.5 - 21.42 ) / 2.8152
Z = 0.03

P ( X < 21.5 ) = P ( Z < 0.03 )
P ( X < 21.5 ) = 0.512

Part c)

P ( X >= 23 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 23 - 0.5 ) =P ( X > 22.5 )

P ( X > 22.5 ) = 1 - P ( X < 22.5 )
Standardizing the value

Z = ( 22.5 - 21.42 ) / 2.8152
Z = 0.38

P ( Z > 0.38 )
P ( X > 22.5 ) = 1 - P ( Z < 0.38 )
P ( X > 22.5 ) = 1 - 0.648
P ( X > 22.5 ) = 0.352

Part d)

P ( 21 <= X <= 26 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 21 - 0.5 < X < 26 + 0.5 ) = P ( 20.5 < X < 26.5 )

P ( 20.5 < X < 26.5 )
Standardizing the value

Z = ( 20.5 - 21.42 ) / 2.8152
Z = -0.33
Z = ( 26.5 - 21.42 ) / 2.8152
Z = 1.8
P ( -0.33 < Z < 1.8 )
P ( 20.5 < X < 26.5 ) = P ( Z < 1.8 ) - P ( Z < -0.33 )
P ( 20.5 < X < 26.5 ) = 0.9644 - 0.3719
P ( 20.5 < X < 26.5 ) = 0.5925