Assume your findings are normally distributed. Using your individual mean (3799.

Question

Assume your findings are normally distributed. Using your individual mean (3799.25), (actual) standard deviation (3303.72), and your z-table, find the following probabilities. Be sure to include a graph of the curve with data elements displaying the correct shading of your area(s): A. a response less than the 15th element of your dataset. B. A response between 15th and 25th element of your dataset. C. A response less than the 15th element or greater than the 25th Element of your dataset.

220 383 450 453 817 1080 1148 1200 1250 1300 1359 1511 1515 1587 1667 1800 1917 1920 2000 2262 2278 2794 2906 3010 3911 3928 4118 4256 5108 6200 6500 6816 6862 7365 8000 8907 9797 10175 11200 12000

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 3799.25
standard Deviation ( sd )= 3303.72
PART A.
P(X < 15) = (15-3799.25)/3303.72
= -3784.25/3303.72= -1.1455
= P ( Z <-1.1455) From Standard Normal Table
= 0.126

PART B.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 15) = (15-3799.25)/3303.72
= -3784.25/3303.72 = -1.1455
= P ( Z <-1.1455) From Standard Normal Table
= 0.126
P(X < 25) = (25-3799.25)/3303.72
= -3774.25/3303.72 = -1.1424
= P ( Z <-1.1424) From Standard Normal Table
= 0.1266
P(15 < X < 25) = 0.1266-0.126 = 0.0006

PART C.
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 15) = (15-3799.25)/3303.72
= -3784.25/3303.72= -1.1455
= P ( Z <-1.1455) From Standard Normal Table
= 0.126
P(X > 25) = (25-3799.25)/3303.72
= -3774.25/3303.72 = -1.1424
= P ( Z >-1.1424) From Standard Normal Table
= 0.8734
P( X < 15 OR X > 25) = 0.126+0.8734 = 0.9994

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