Question

74% of all Americans live in cities with population greater than 100,000 people. If 39 Americans are randomly selected, find the probability that

a. Exactly 27 of them live in cities with population greater than 100,000 people.

b. At most 27 of them live in cities with population greater than 100,000 people.

c. At least 30 of them live in cities with population greater than 100,000 people.

d. Between 28 and 32 (including 28 and 32) of them live in cities with population greater than 100,000 people.

Answer 1

Mean = n * P = ( 39 * 0.74 ) = 28.86
Variance = n * P * Q = ( 39 * 0.74 * 0.26 ) = 7.5036
Standard deviation = = 2.7393

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 - 28.86 ) / 2.7393
Z = -0.86
Z = ( 27.5 - 28.86 ) / 2.7393
Z = -0.5
P ( -0.86 < Z < -0.5 )
P ( 26.5 < X < 27.5 ) = P ( Z < -0.5 ) - P ( Z < -0.86 )
P ( 26.5 < X < 27.5 ) = 0.3098 - 0.1945
P ( 26.5 < X < 27.5 ) = 0.1153

Part b)

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

P ( X < 27.5 )
Standardizing the value

Z = ( 27.5 - 28.86 ) / 2.7393
Z = -0.5

P ( X < 27.5 ) = P ( Z < -0.5 )
P ( X < 27.5 ) = 0.3085

Part c)

P ( X >= 30 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 30 - 0.5 ) =P ( X > 29.5 )

P ( X > 29.5 ) = 1 - P ( X < 29.5 )
Standardizing the value

Z = ( 29.5 - 28.86 ) / 2.7393
Z = 0.23

P ( Z > 0.23 )
P ( X > 29.5 ) = 1 - P ( Z < 0.23 )
P ( X > 29.5 ) = 1 - 0.591
P ( X > 29.5 ) = 0.409

Part d)

P ( 28 <= X <= 32 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 28 - 0.5 < X < 32 + 0.5 ) = P ( 27.5 < X < 32.5 )

P ( 27.5 < X < 32.5 )
Standardizing the value

Z = ( 27.5 - 28.86 ) / 2.7393
Z = -0.5
Z = ( 32.5 - 28.86 ) / 2.7393
Z = 1.33
P ( -0.5 < Z < 1.33 )
P ( 27.5 < X < 32.5 ) = P ( Z < 1.33 ) - P ( Z < -0.5 )
P ( 27.5 < X < 32.5 ) = 0.908 - 0.3098
P ( 27.5 < X < 32.5 ) = 0.5983