Answered Essay: 1. With a mean of 875 and a standard deviation of 125, what is the likelihood that a randomly selected observation

1. With a mean of 875 and a standard deviation of 125, what is the likelihood that a randomly selected observation will lie in the following ranges? Show calculation

a. More than 875

b. Between 750 and 1200

c. Less than 975

d. Between 750 and 1100

e. More than 875

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)

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