Question

according to a study done by Oxford research Center, 39% of adult americans believe that relationships is now out of date. Suppose a random sample of 500 adult english individuals is asked whether relationships is out of date.

A.) what is probability that in a random sample of 500 adult english individauls at most 200 believe that relationships is out of date?

B.) What is the probability that in a random sample of 500 adult english individuals between 40% and 45% believe that relationships is out of date?

C.) Would it be unusual for a random sample of 500 adult english individuals to result in 210 or more who believe that relationships is out of date?

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 500 * 0.39 ) = 195
Variance = n * P * Q = ( 500 * 0.39 * 0.61 ) = 118.95
Standard deviation = √(variance) = √(118.95) = 10.9064

P ( X <= 200 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 200 + 0.5 ) = P ( X < 200.5 )

X ~ N ( µ = 195 , σ = 10.9064 )
P ( X < 200.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 200.5 - 195 ) / 10.9064
Z = 0.5
P ( ( X - µ ) / σ ) < ( 200.5 - 195 ) / 10.9064 )
P ( X < 200.5 ) = P ( Z < 0.5 )
P ( X < 200.5 ) = 0.6915

Part b)

Sampling distribution of p̂ is approximately normal if np >=10 and n (1-p) >= 10
n * p = 500 * 0.39 = 195
n * (1 - p ) = 500 * (1 - 0.39) = 305
Mean = = p = 0.39
Standard deviation = = 0.021813

X ~ N ( µ = 0.39 , σ = 0.021813 )
P ( 0.4 < X < 0.45 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 0.4 - 0.39 ) / 0.021813
Z = 0.46
Z = ( 0.45 - 0.39 ) / 0.021813
Z = 2.75
P ( 0.46 < Z < 2.75 )
P ( 0.4 < X < 0.45 ) = P ( Z < 2.75 ) - P ( Z < 0.46 )
P ( 0.4 < X < 0.45 ) = 0.997 - 0.6772
P ( 0.4 < X < 0.45 ) = 0.3198

Part c)

P ( X >= 210 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 210 - 0.5 ) =P ( X > 209.5 )

X ~ N ( µ = 195 , σ = 10.9064 )
P ( X > 209.5 ) = 1 - P ( X < 209.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 209.5 - 195 ) / 10.9064
Z = 1.33
P ( ( X - µ ) / σ ) > ( 209.5 - 195 ) / 10.9064 )
P ( Z > 1.33 )
P ( X > 209.5 ) = 1 - P ( Z < 1.33 )
P ( X > 209.5 ) = 1 - 0.9082
P ( X > 209.5 ) = 0.0918

It is not unusual, because probability is more than 0.05.