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```North Carolina State University
STAT 370: Probabilityy and Statistics for
Engineers
[Section 002]
Instructor: Hua Zhou
Harrelson Hall 210
11:45AM-1:00PM, Apr 11, 2012
Midterm 2
Announcements
• Midterm 2 returned
• Quiz 4 returned
• HW 9 (continuous R.V., 10pt) due Friday Apr 13 @
11:59PM
• HW 10 (normal distribution, 10pt) due Friday Apr 13
@ 11:59PM
• HW 11 (sampling dist and C.I., 10pt) due Apr 20 @
11:59PM
• HW 12 (hypothesis testing,
testing 10pt) due Apr 27 @
11:59PM
Plan
• Last time:
Continuous random variable, normal distribution,
standard normal table
• Today:
Normal probability calculation, start the last topic
(inferential statistics)
1
The Standard Normal Table
• The standard normal table is a table of areas under the
standard normal density curve. The table entry for each
value z is the area under the curve to the left of z. Normal
table gives us CDF values (P(Z<=z))
Standard Normal Distribution
• Here are some examples of probability calculations from
N(0,1):
The 68-95-99.7 Rule for Normal distribution

Pr[  1  Z  1]  .6826
Pr[  2  Z  2]  .9546
P r[  3  Z  3]  .9974
This is where the 68-95-99.7 empirical rule comes from.

2
Means/Variances Under Linear Transformation
Non-standard Normal variables: calculation of Pr(a<X<b)
• Consider X~ N(µ,σ2), how to calculate probability
involving X?
• Question: What is the distribution of Z= (X-µ)/σ ?
Answer: Z is a standard normal.
This is because:
1) Linear transformations applied to normal variables
preserve normality (e.g. Y= aX+b, then if X is normal so
i Y)
is
Y).
2) Remember common rules for means and variances:
• If X is a RV and a and b are constants, then
1.
2
3.
E ( X  a)  E ( X )  a
E (aX )  aE ( X )
E (aX  b) aE ( X )  b
Var ( X  a)  Var ( X )
1.
2.
Var (aX )  a 2Var ( X )
3
Var (aX  b)  a 2Var ( X )
Non-standard Normal variables: calculation of Pr(a<X<b)
Example:
1. Assume X is normal with µ=30
µ 30 and σ2=25.
25. Determine
Pr[X <= 35]:
Solution:
i. Convert x-value to a z-score
xvalue  mean 35  30

=1.00 = z-score
std .deviation
5
Thi tells
This
t ll us how
h
many standard
t d d deviations
d i ti
((and
d which
hi h
direction) away from the mean the x-value is.
ii. P[X<= 35] = Pr[Z <= 1.00] = 0.8413 [using the Tablle]
In class exercise
• Let X  N(6,5. 2 ) . Find:
• a)
P(6  X  12)  P( X  12)  P( X  6)
 .....
• b)
P( X  21)  1  P( X  21)  ......
• c)
P( X  6  5)  P(5  X  6  5)  .....
3
In class exercise [solutions]
Example: Young Women’s Height
• The heights of young women are approximately normal with
mean = 64.5 inches and std.dev. = 2.5 inches.
• Let X  N(6,5. 2 ) . Find:
• a) P(6  X  12)  P( X  12)  P( X  6)
12  6
66
)  P(Z 
)  P(Z  6/ 5)  P(Z  0)
5
5
 0.8849  0.5  0.3849
 P(Z 
• b)
P( X  21)  1  P( X  21)  1  P(Z 
21  6
)
5
 1  P(Z  3)  1  .9986  .0014
• c)
P( X  6  5)  P(5  X  6  5)  P(1  X  11)
 P( X  11)  P( X  1)  P(Z  1)  P(Z  1)
 0.8413  0.1586  0.6827
Example: Young Women’s Height (cont’d)
The heights of young women are approximately normal
with mean = 64.5 inches and std.dev. = 2.5 inches.
(a) Determine the % of young women between 62 and 67 ?
(b) Determine the % of young women lower than 62 or
taller than 67?
(c) Determine the % between 59.5 and 62?
(d) Determine the % taller than 68.25?
14
In class exercise
• SAT verbal scores are known to be normally distributed
with mean µ = 505 and standard deviation σ = 110 based
on data from the College Board.
(a) Draw a normal curve with the parameters labeled.
(b) Shade the region that represents the proportion of test
takers who scored less than 395.
(c) The area under the normal curve to the left of X=395
is 0
0.1587.
1587 Provide two interpretations of this result
result. (The
probability that a randomly selected person scores less
than 395, and the proportion of the population that scores
less than 395.)
(d) Determine the 95th percentile of this distribution. Begin
with 1.645 from Z.
4
In class exercise (cont’d)
• SAT verbal scores are known to be normally distributed
with mean µ = 505 and standard deviation σ = 110 based
on data from the College Board.
(e) Determine the values that determine the middle 95%
of the scores (by middle we refer to an interval that is
In class exercise
The heights of a pediatrician’s 200 three-year-old females
are normally distributed with mean 38.72 inches and
standard deviation 3.17 inches. The pediatrician wishes
to determine the middle 98% of heights (thus we want the
1st and 99th percentile). What are these values?
In class exercise
The diameters of bearing journals ground on a particular
grinder can be described as normally distributed with mean
2 005 in.
2.005
in and standard deviation 0
0.005
005 in.
in
a) If engineering specifications on these diameters are
2.000in.  0.005 in., what fraction of these journals are in
specifications?
(b) Assume that journal diameters are independent of other
journals Four journals are chosen at random
journals.
random, and let Y be
the number of these journals that are within specifications.
Determine the probability mass function of Y, and the
probability that at least one of the four journals is within
specifications.
5
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