Question

69% of all students at a college still need to take another math class. If 41 students are randomly selected, find the probability that a. Exactly 30 of them need to take another math class. b. At most 27 of them need to take another math class. c. At least 29 of them need to take another math class. d. Between 23 and 29 (including 23 and 29) of them need to take another math class.

Answer 1

Mean = n * P = ( 41 * 0.69 ) = 28.29
Variance = n * P * Q = ( 41 * 0.69 * 0.31 ) = 8.7699
Standard deviation = = 2.9614

Part a)

P ( X = 30 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 30 - 0.5 < X < 30 + 0.5 ) = P ( 29.5 < X < 30.5 )

P ( 29.5 < X < 30.5 )
Standardizing the value

Z = ( 29.5 - 28.29 ) / 2.9614
Z = 0.41
Z = ( 30.5 - 28.29 ) / 2.9614
Z = 0.75
P ( 0.41 < Z < 0.75 )
P ( 29.5 < X < 30.5 ) = P ( Z < 0.75 ) - P ( Z < 0.41 )
P ( 29.5 < X < 30.5 ) = 0.7722 - 0.6586
P ( 29.5 < X < 30.5 ) = 0.1137

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 - 28.29 ) / 2.9614
Z = -0.27

P ( X < 27.5 ) = P ( Z < -0.27 )
P ( X < 27.5 ) = 0.3936

Part c)

P ( X >= 29 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 29 - 0.5 ) =P ( X > 28.5 )

P ( X > 28.5 ) = 1 - P ( X < 28.5 )
Standardizing the value

Z = ( 28.5 - 28.29 ) / 2.9614
Z = 0.07

P ( Z > 0.07 )
P ( X > 28.5 ) = 1 - P ( Z < 0.07 )
P ( X > 28.5 ) = 1 - 0.5279
P ( X > 28.5 ) = 0.4721

Part d)

P ( 23 <= X <= 29 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 23 - 0.5 < X < 29 + 0.5 ) = P ( 22.5 < X < 29.5 )

P ( 22.5 < X < 29.5 )
Standardizing the value

Z = ( 22.5 - 28.29 ) / 2.9614
Z = -1.96
Z = ( 29.5 - 28.29 ) / 2.9614
Z = 0.41
P ( -1.96 < Z < 0.41 )
P ( 22.5 < X < 29.5 ) = P ( Z < 0.41 ) - P ( Z < -1.96 )
P ( 22.5 < X < 29.5 ) = 0.6586 - 0.0253
P ( 22.5 < X < 29.5 ) = 0.6333