Question

An investment company knows that the rates of return on its portfolios have a mean of 7.45 percent, with a standard deviation of 3.82 percent. The company selects a sample of 144 portfolios to analyze. Assume the company has tens of thousands of portfolios. (Careful- "percent" is just a unit here!) You do NOT need to check CLT here.

A. What is the probability that the mean of the sample is smaller than 7 percent?

B. What is the probability that the mean of the sample is larger than 7.3 percent?

C. What is the probability that the mean of the sample is between 7.5 and 8.2 percent?

Answer 1

Part A)

X ~ N ( µ = 7.45 , σ = 3.82 )
P ( X < 7 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 7 - 7.45 ) / ( 3.82 / √144 )
Z = -1.41
P ( ( X - µ ) / ( σ/√(n)) < ( 7 - 7.45 ) / ( 3.82 / √(144) )
P ( X < 7 ) = P ( Z < -1.41 )
P ( X < 7 ) = 0.0787

Part B)
X ~ N ( µ = 7.45 , σ = 3.82 )
P ( X > 7.3 ) = 1 - P ( X < 7.3 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 7.3 - 7.45 ) / ( 3.82 / √ ( 144 ) )
Z = -0.47
P ( ( X - µ ) / ( σ / √ (n)) > ( 7.3 - 7.45 ) / ( 3.82 / √(144) )
P ( Z > -0.47 )
P ( X > 7.3 ) = 1 - P ( Z < -0.47 )
P ( X > 7.3 ) = 1 - 0.3187
P ( X > 7.3 ) = 0.6813

Part C)
X ~ N ( µ = 7.45 , σ = 3.82 )
P ( 7.5 < X < 8.2 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 7.5 - 7.45 ) / ( 3.82 / √(144))
Z = 0.16
Z = ( 8.2 - 7.45 ) / ( 3.82 / √(144))
Z = 2.36
P ( 0.16 < Z < 2.36 )
P ( 7.5 < X < 8.2 ) = P ( Z < 2.36 ) - P ( Z < 0.16 )
P ( 7.5 < X < 8.2 ) = 0.9908 - 0.5624
P ( 7.5 < X < 8.2 ) = 0.4284