Question

75% of adults only speak english. a sample of 150 adults are selected. let the random variable X represent the number who speaks only english.

describe the probability distribution of X (the type and its 2 parameters)

calculate p(x=115)

calculate p(X<110)

calculate p(X>120)

Answer 1

Condition check for Normal Approximation to Binomial
n * P >= 10 = 150 * 0.75 = 112.5
n * (1 - P ) >= 10 = 150 * ( 1 - 0.75 ) = 37.5

Using Normal Approximation to Binomial
Mean = n * P = ( 150 * 0.75 ) = 112.5
Variance = n * P * Q = ( 150 * 0.75 * 0.25 ) = 28.125
Standard deviation = √(variance) = √(28.125) = 5.3033

Part a)

P ( X = 115 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 115 - 0.5 < X < 115 + 0.5 ) = P ( 114.5 < X < 115.5 )

X ~ N ( µ = 112.5 , σ = 5.3033 )
P ( 114.5 < X < 115.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 114.5 - 112.5 ) / 5.3033
Z = 0.38
Z = ( 115.5 - 112.5 ) / 5.3033
Z = 0.57
P ( 0.38 < Z < 0.57 )
P ( 114.5 < X < 115.5 ) = P ( Z < 0.57 ) - P ( Z < 0.38 )
P ( 114.5 < X < 115.5 ) = 0.7157 - 0.648
P ( 114.5 < X < 115.5 ) = 0.0676

Part b)

P ( X < 110 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 110 - 0.5 ) = P ( X < 109.5 )

X ~ N ( µ = 112.5 , σ = 5.3033 )
P ( X < 109.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 109.5 - 112.5 ) / 5.3033
Z = -0.57
P ( ( X - µ ) / σ ) < ( 109.5 - 112.5 ) / 5.3033 )
P ( X < 109.5 ) = P ( Z < -0.57 )
P ( X < 109.5 ) = 0.2843

Part c)

P ( X > 120 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 120 + 0.5 ) = P ( X > 120.5 )

X ~ N ( µ = 112.5 , σ = 5.3033 )
P ( X > 120.5 ) = 1 - P ( X < 120.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 120.5 - 112.5 ) / 5.3033
Z = 1.51
P ( ( X - µ ) / σ ) > ( 120.5 - 112.5 ) / 5.3033 )
P ( Z > 1.51 )
P ( X > 120.5 ) = 1 - P ( Z < 1.51 )
P ( X > 120.5 ) = 1 - 0.9345
P ( X > 120.5 ) = 0.0655