WORKSHEET-NORMAL DESTRIBUTIONS . For each probles below draw a piecture of the n
ID: 3317617 • Letter: W
Question
WORKSHEET-NORMAL DESTRIBUTIONS . For each probles below draw a piecture of the normal curve ansd shade the area you bave to find Di sa Let Z epresent a variable following a standard noemal distribvation. (a) Find the proportion that is less than2.00 (b) Find the proportice that is betwors z-13 and-1.75 (c) Fiad the proportion that is greater than 1.86 (d) Find the z-score for the 64th percemtile (e) Find the z-scores that bound the middle 50% of all data ( Find the -score for the 24th peroentile. 2. Former ISU basketball player Kelvin Cato is 83 inches tall. Assuming that heights follow approxi- mately a normal distribution with mean 70 and standard deviation (a) what is his corresponding z-score? 3, (b) what proportion of men are taller than him?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 ) = 0
standard Deviation ( sd )= 1
a.
P(X < -2) = (-2-0)/1
= -2/1= -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
b.
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -0.13) = (-0.13-0)/1
= -0.13/1 = -0.13
= P ( Z <-0.13) From Standard Normal Table
= 0.4483
P(X < 1.75) = (1.75-0)/1
= 1.75/1 = 1.75
= P ( Z <1.75) From Standard Normal Table
= 0.9599
P(-0.13 < X < 1.75) = 0.9599-0.4483 = 0.5117
c.
P(X > 1.86) = (1.86-0)/1
= 1.86/1 = 1.86
= P ( Z >1.86) From Standard Normal Table
= 0.0314
d.
P ( Z < x ) = 0.64
Value of z to the cumulative probability of 0.64 from normal table is 0.358459
P( x-u/s.d < x - 0/1 ) = 0.64
That is, ( x - 0/1 ) = 0.358459
--> x = 0.358459 * 1 + 0 = 0.358459
e.
LOWER/BELOW
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.67449
P( x-u/s.d < x - 0/1 ) = 0.25
That is, ( x - 0/1 ) = -0.67449
--> x = -0.67449 * 1 + 0 = -0.67449
UPPER/TOP
P ( Z > x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is 0.67449
P( x-u / (s.d) > x - 0/1) = 0.25
That is, ( x - 0/1) = 0.67449
--> x = 0.67449 * 1+0 = 0.67449
f.
P ( Z < x ) = 0.24
Value of z to the cumulative probability of 0.24 from normal table is -0.706303
P( x-u/s.d < x - 0/1 ) = 0.24
That is, ( x - 0/1 ) = -0.706303
--> x = -0.706303 * 1 + 0 = -0.706303
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