uiuc 3.3 – normal distribution€¦ · 3.3 – normal distribution stepanov dalpiaz nguyenrv µ...
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STAT 400 UIUC 3.3 – Normal Distribution Stepanov
Dalpiaz Nguyen
µ – mean s – standard deviation
,
- ¥ < x < ¥.
Standard Normal Distribution.
mean
0
standard deviation
1
N ( 0 , 1 )
÷øöç
èæ 2 , σμ N
( ) ( ) 2 2
σμ 2
2 1σπ
--= xexf
( Gaussian)
-
continuous RV
parameters - defininga specie
belt - shaped distributioncurve
X -
=
2
Example: For the standard normal distribution, find the area to the left of
z = 1.24
Area to the left of z = 1.24 is 0.8925.
Z ~ N ( 0 , 1 )
X ~ N ( µ , s 2 )
X = µ + s Z
σμX Z -=
T standard normal
→ 2- = 1.04
2- = O.96
O
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Example 1: At Initech , the salaries of the employees are normally distributed with mean µ = $36,000 and standard deviation s = $5,000. a) Mr. Smith is paid $42,000. What proportion of the employees of Initech are paid less that Mr. Smith? b) What proportion of the employees have their salaries over $40,000? c) Suppose 10 Initech employees are randomly and independently selected. What is the probability that 3 of them have their salaries over $40,000? d) What proportion of the employees have their salaries between $30,000 and $40,000?
Let X -- salaries of employees at Initech .
X ~ N ( µ = 36000, 62 = 50002£ 6=5000
I ( X C 42,000) = I ( XII g 42000%0000-1=112-41.2)F = 0.88490
I ( x > 40, ooo ) = I ( XII > 40000300g )w
zu N (O, I)= I ( Z> 0.8) = I - I ( Z S 0.8) = I - 0.7881 = 0.21190
Y = # of employees that have salaries over $40,000.
Y ~ Binom ( n -- 10, p = I (X > 40,000)= 0.2119)
I ( Y =3) = ( t} ) O .21193 ( l - 0.2119)
"- 0.21560
I (30,0004×440,000)↳X - µ 40000 - 36000= I (3¥00 <g- LE)
sooow
Z n NCO, I)= If -1.2C Z C 0.8 ) #¥.#= I ( 2- ( 0.8 ) - I ( Z C - 1.2)= 0. 7881 - O
.1151 = 0.67300
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e) Mrs. Jones claims that her salary is high enough to just put her among the highest paid 15% of all employees working at Initech . Find her salary. f) Ms. Green claims that her salary is so low that 90% of the employees make more than she does. Find her salary. Example 2: Suppose that the lifetime of Outlast batteries is normally distributed with mean µ = 240 hours and unknown standard deviation. Suppose also that 20% of the batteries last less than 219 hours. Find the standard deviation of the distribution of the lifetimes.
Let m = Mrs.
Torres'
salary .
I ( X > m ) = 0.15 ⇐ I ( X ( m) = I- 0.15=0.85
⇒ efx-ftsm-soooIJ-0-85m-soooo-i.ae⇒ I ( IT m}#W) = 0.85 ⇒ 5000
⇒ m--41,2000
= 1.04
Let m = Ms.Green's salary .
I ( X 7, m) = 0.9 ⇒ I ( X (m ) = I - 0.9=0.1"
⇒ If enjoy) -- o. ,⇒ I Czc m-I6§) = O
. , ⇒ m-3GW°_ = - 1.295000
- ⇒ m-- 29,6000
z = - 1.29
Let X -
- lifetime of Outlast batteries .
X N N ( µ -- 240, 6- ? )I ( XS 219) -- 0.2
⇒ e f z (219-346)=0.2-
Z= - 0.84
So, 219-62402=0.2 ⇒ 6=25haer①
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Let X be normally distributed with mean µ and standard deviation s. Then the moment-generating function of X is
M X ( t ) = .
Let Y = a X + b. Then M Y ( t ) = e b t M X ( a t ).
Therefore, Y is normally distributed with mean a µ + b and variance a 2 s 2 ( standard deviation | a | s ). Example 1: (cont.) g) All Initech employees receive a memo instructing them to put away 4% of their salaries plus $100 per month ( $1,200 per year ) in a special savings account to supplement Social Security. What proportion of the employees would put away more than $3,000 per year?
2 σ μ 22 tte +in
Y = 0.04 X t 1200
So, ply = 0.04 pext 1200=2640
by = 0.04 Mx = 200.
⇒ Y N N ( pre -- 2640 , 6=200)I ( Y > 30007 = I ( z > 30002%640-1=1 ( Z > 1.8)
= I - 0.9641 = 0.03590