2 Standard Normal CDF F(z) 1.0000 0.9382 -- 0.6103-- 0.2327 a) b) c3 Z Using the

Question

2 Standard Normal CDF F(z) 1.0000 0.9382 -- 0.6103-- 0.2327 a) b) c3 Z Using the CDF of a standard normal variable (Z) above, find the values for: a) b) d) what is the probability that a randomly selected z falls between a) and e)? e) The probability isthat a randomly selected z falls between 0 and b). f) What is the estimated average height of the PDF for this distributino between -2.99 and -3.00? (you don't have to be very exact, but you should explain). 8) What is the average height of the PDF between Z-0 and Z-b)? (to 4 decimal places)

Explanation / Answer

Question 2

We can see part (a) to (c) from Z - table directly

(a) F-1(0.2327) = -0.73

(b) F-1(0.6103) = 0.28

(c) F-1(0.9382) = 1.54

(d) Pr(a < z < c) = 0.9382 - 0.2327 = 0.7055

(e) Pr(0 < z < b) = 0.6103 - 0.5 = 0.1103

(f) Average height of the pdf between Z = -2.99 to Z = -3.00

Here as the values are very near we know that derivative of cumulative probability distribution of CDF is pdf.

As Area between curve = Height * Distance (and we can take it is an averagE)

so Area between the given points = Pr(-2.99 < Z < -3.00) = 0.001395 - 0.00135 = 0.000045

0.000045 = Height * (3 - 2.99)

Height of the PDF = 0.000045/0.01 = 0.0045

(g) Here the average height of PDF between Z = 0 to b

Area = HEight * (b-0)

0.1103 = Average Height * 0.28

Average Height = 0.1103/0.28 = 0.3939

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