Question

A population has a mean age of 50 years and a standard deviation equal to 10 years. To study the population, you decide to sample 36 individuals and measure their age. What is the probability that a sample mean age is between 47.5 and 52.5 years?

Answer

What is the probability of selecting a sample of 36 individuals that has a sample mean of 55 years or more?

Answer

What is the probability of selecting a sample of 36 individuals that has a sample mean of 50 years or less?

Answer
  

What assumption did you make to calculate the probabilities? Answer

Why can you make your assumption? Answer

Answer 1

a)

X ~ N ( µ = 50 , σ = 10 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 47.5 - 50 ) / ( 10 / √(36))
Z = -1.5
Z = ( 52.5 - 50 ) / ( 10 / √(36))
Z = 1.5
P ( 47.5 < X̅ < 52.5 ) = P ( Z < 1.5 ) - P ( Z < -1.5 )
P ( 47.5 < X̅ < 52.5 ) = 0.9332 - 0.0668 (from Z table)
P ( 47.5 < X̅ < 52.5 ) = 0.8664

b)

X ~ N ( µ = 50 , σ = 10 )
P ( X > 55 ) = 1 - P ( X < 55 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 55 - 50 ) / ( 10 / √ ( 36 ) )
Z = 3
P ( ( X - µ ) / ( σ / √ (n)) > ( 55 - 50 ) / ( 10 / √(36) )
P ( Z > 3 )
P ( X̅ > 55 ) = 1 - P ( Z < 3 )
P ( X̅ > 55 ) = 1 - 0.9987 (from Z table)
P ( X̅ > 55 ) = 0.0013

c)

X ~ N ( µ = 50 , σ = 10 )
P ( X < 50 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 50 - 50 ) / ( 10 / √36 )
Z = 0
P ( ( X - µ ) / ( σ/√(n)) = ( 50 - 50 ) / ( 10 / √(36) )
P ( X < 50 ) = P ( Z < 0 )
P ( X̅ < 50 ) = 0.5

d)

We assume that distribution is approximately normally distributed.