Question & Answer: 1. With a mean of 875 and a standard deviation of 125, what is the likelihood that a randomly selected observation…

Expert Answer

(a). x = 875

Z = (x-μ)/σ = (875-875)/125 = 0

P(x>875) = 1 -NORMDIST(0) = 1 – 0.5 = 0.5   (Alternative, we can lookup for z value 0 in the standard normal table)

Likelihood that a randomly selected observation will be more than 875 = 0.5 or 50%

(b). For x=750, z = (750-875)/125 = -1

For x=1200, z = (1200-875)/125 = 2.6

Corresponding to z=-1, cumultive distribution function = NORMSDIST(-1) = 0.1587

Corresponding to z=2.6, cumultive distribution function = NORMSDIST(2.6) = 0.9953

Likelihood that a randomly selected observation will lie between 750 and 1200 = 0.9953 – 0.1587 = 0.8367 or 83.67%

(c) For x=975, z = (975-875)/125 = 0.8

Corresponding to z=0.8, cumultive distribution function = NORMSDIST(0.8) = 0.7881

Likelihood that a randomly selected observation will be less than 975 = 0.7881 or 78.81%

(d). For x=750, z = (750-875)/125 = -1

For x=1100, z = (1100-875)/125 = 1.8

Corresponding to z=-1, cumultive distribution function = NORMSDIST(-1) = 0.1587

Corresponding to z=1.8, cumultive distribution function = NORMSDIST(1.8) = 0.9641

Likelihood that a randomly selected observation will lie between 750 and 1100 = 0.9641 – 0.1587 = 0.8054 or 80.54%

(e) This is same as (a)