Distribution of Values

Question

A distribution of values is normal with a mean of 729.2 and a standard deviation of 30.8.

Find the probability that a randomly selected value is between 662.1 and 826.5

Enter your answer as a number accurate to 4 decimal places. * Note all z scores must be rounded to the near tee hundredth

Answer 1

Mariana,

You can start by declaring your variables:

Population Mean μ = 729.2

Population SD σ = 30.8

Random Variables X₁ and X₂ = 662.1 and 826.5, respectively.

Do a quick-check guide with Chebychev's rule for area under the curve (AUC), this helps to verify our answer later:

~68% of AUC = μ + σ = [ 698.4 , 760 ]

~95% of AUC = μ + 2σ = [ 667.6 , 790.8 ] -> X₁ is in this range

~99% of AUC = μ + 3σ = [ 636.8, 821.6 ] -> X₂ is beyond this range

1. This tells you that the final answer should be very large, and nearly all of the AUC.

Z_x1 = (X₁ - μ)/σ = (662.1 - 729.2)/ 30.8 = -2.18 (Truncated to nearest .01 for Z-value lookup)

1. Left-Side Z-Look Up (AUC for values of z smaller than -2.18) = 0.0146

Z_x2 = (X₂ - μ)/σ = (826.5 - 729.2)/ 30.8 = 3.16 (Truncated to nearest .01 for Z-value lookup)

1. Right-Side Z-Look Up (AUC for values of z smaller than 3.16) = .9992

Area between is the difference of the two values (Right_Side - Left_Side), or 0.9992 - 0.0146 = 0.9846 or 98.46%