Question

1. The weights of adults (in kg) follows a normal distribution with a mean of 67 and a standard deviation of 11. For a random sample of 64 adults, find the probability that the mean weight of the sample is at most 63 kg.

2. Suppose that 50% of politicians are lawyers. Find the probability that of a random sample of 400 politicians, at least 47% are lawyers.

Answer 1

Part 1)

X ~ N ( µ = 67 , σ = 11 )
P ( X <= 63 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 63 - 67 ) / ( 11 / √64 )
Z = -2.9091
P ( ( X - µ ) / ( σ/√(n)) < ( 63 - 67 ) / ( 11 / √(64) )
P ( X <= 63 ) = P ( Z < -2.91 )
P ( X̅ <= 63 ) = 0.0018

part 2)

47% of 400 = 188

P ( X >= 188 )

Using Normal Approximation to Binomial
Mean = n * P = ( 400 * 0.5 ) = 200
Variance = n * P * Q = ( 400 * 0.5 * 0.5 ) = 100
Standard deviation = √(variance) = √(100) = 10

P ( X >= 188 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 188 - 0.5 ) =P ( X > 187.5 )

X ~ N ( µ = 200 , σ = 10 )
P ( X > 187.5 ) = 1 - P ( X < 187.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 187.5 - 200 ) / 10
Z = -1.25
P ( ( X - µ ) / σ ) > ( 187.5 - 200 ) / 10 )
P ( Z > -1.25 )
P ( X > 187.5 ) = 1 - P ( Z < -1.25 )
P ( X > 187.5 ) = 1 - 0.1056
P ( X > 187.5 ) = 0.8944