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F7 Mathematics and Statistics
Chapter 16 The Normal Distribution and Its Applications
F7-MS-Ch16 - 1
16 The normal distribution (正態分佈)
§ 16.1
The normal distribution 正態分佈
It is a continuous probability distribution (連續概率分佈) with mean  and variance  2 (or
standard deviation  ) such that its p.d.f. f (x) is given by f ( x) 
1
2

( x )2
e
2 2
for all values
of x .
If x is distributed in this way, we write x ~ N (  ,  2 )
§ 16.2
Properties of normal distribution 正態分佈的特性
(a) The mean (平均數), median (中位數) and the mode (眾數) are equal.
(b) The curve is bell-shaped (鐘形) and symmetric (對稱) about the mean.
(c) The curve approaches (趨近) the x-axis (or z-axis) asymptotically (漸近地).
(d)
The flatness (平坦程度) of the curve is determined by  and the position (位置) the
curve is determined by  .
(e)
The total area under the normal curve is 1.
(f)
The area under the curve from x = a to x = b represents the proportion (probability) of
data (數據) for x takes up values from a to b, or in symbolic notation P( a  x  b )
(g)
The probability that x takes up a single value (某一數值) is zero (零). When x is
continues.
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F7 Mathematics and Statistics
Chapter 16 The Normal Distribution and Its Applications
(h)
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The probability that x takes up a given standard deviation within the mean is a constant
(在平均值內某指定標準差一常數).
Suppose x ~ N(  ,  2 ), then
(i) P(     x     ) =0.6826
(ii)
P(   2  x    2 ) =0.9544
(iii)
P(   3  x    3 ) =0.9974
§ 16.3
Application of normal distribution 正態分佈的應用
Common application of normal distribution are
(a)
height
(b)
mass
(c)
examination results
(d)
I.Q. score
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Chapter 16 The Normal Distribution and Its Applications
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§ 16.3.1 The standard normal distribution 標準正態分佈
If x ~ N (0,1), then x is a standard normal variable (標準正態變量) or x follows (依循) a standard
normal distribution. The p.d.f. f(x) of standard normal distribution is f(x)=
§ 16.3.2 Standard normal table and standardization
(a)
1
2
e

x2
2
標準正態分佈表與標準化
Standard normal table 標準正態分佈表
If z ~ N (0, 1), then the probability that z lies within the range from 0 to k, is given by
P( 0  z  k ) = area under the curve from x = 0 to x = k, and the value may be easily found
from the standard normal table.
(b) Standardization 標準化
If x ~ N(  ,  2 ), then we may standardize x to a standard normal variable z by letting z 
then z ~ N(0,1).
Example 1
If z is a standard normal variable (標準正態變量),evaluate the following (計算下列):
(a) P( 0  z  1.2 )
= 0.3849 (to 4 d.p)
F7-MS-Ch16 - 3
x

,
F7 Mathematics and Statistics
Chapter 16 The Normal Distribution and Its Applications
(b) P( 0  z  2.54)
= 0.4945 (to 4 d.p)
(c) P(  0.5  z  0 )
= 0.1915 (to 4 d.p)
(d) P(  1.39  z  0 )
= 0.4177 (to 4 d.p)
(e) P( 1.1  z  1.78 )
= 0.4625 – 0.3643
= 0.0982
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Chapter 16 The Normal Distribution and Its Applications
(f)
P(  056  z  0.47 )
= 0.2123 + 0.1808
= 0.3931 (to 4 d.p)
(g) P(  1.86  z  0.11 )
= 0.4686 – 0.0438
=0.4248
(h) P(  0.32  z )
= 0.1255 +0.5
=0.6255 (to 4 d.p)
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Chapter 16 The Normal Distribution and Its Applications
(i)
P( z  0.69 )
= 0.5 – P(  0.69  z  0)
= 0.5 – 0.2549
= 0.2451 (to 4 d.p)
(j)
P( z  0.61)
= 0.5 – P( 0  z  0.61)
= 0.5 – 0.2291
= 0.2709
(k) P( z  1.66 )
= 0.5 + 0.4515
= 0.9515
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Chapter 16 The Normal Distribution and Its Applications
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Example 2
If z ~ N(0,1), which of the following values P1,P2,P3,P4 is the greatest (最大)?
(a) P1 = P( 0  z  1)
= 0.3413
(b) P2 = P( 1  z  2 )
= 0.4772 – 0.3413
= 0.1359
(c) P3 = P(  0.5  z  0.5 )
= (0.1915)2
= 0.383
(d) P4 = P(  0.4  z  0.6 )
= 0.1554 + 0.2257
= 0.3811
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F7 Mathematics and Statistics
Chapter 16 The Normal Distribution and Its Applications
Example 3
If z ~N(0,1),find the value of k for each of the following :
(a) P( 0  z  k ) = 0.3289
k 0
 0.95
1
k =0.95
(b) P( 0  z  k ) = 0.4920
k = 2.41
(c) P( k  z  0 ) = 0.3531
–k = 1.05
k = –1.05
(d) P( k  z  0 ) = 0.1064
k = –0.27
(e) P( k  z  0.2 ) = 0.2347
P(k<z<0) = 0.2347 – 0.0793
–k = 0.4
k = –0.4
(f)
P( k  z  1.23 ) = 0.0912
0.3907 – P( 0  z  k ) = 0.0912
P( 0  z  k ) = 0.2995
k = 0.84
(g) P( 1  z  k ) = 0.5536
P( 0  z  k ) = 0.5536 – 0.3413
P( 0  z  k ) = 0.2123
k = 0.556
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Chapter 16 The Normal Distribution and Its Applications
(h) P(-1.4<z<k) = 0.2207
P(k<z<0) = 0.4192 – 0.2207
P(k<z<0) = 0.1985
k = – 0.52
(i)
P(z>k) = 0.0668
P( 0  z  k ) = 0.5 – 0.0668
k = 1.50
(j)
P(z<k) = 0.2514
P(k<z<0) = 0.5 – 0.2514
P(k<z<0) = 0.2486
k = – 0.67
(k) P( z  k ) = 0.9292
P(–k<z<0) = 0.9292 – 0.5
P(–k<z<0) = 0.4292
k = – 1.47
(l)
P(z<k) = 0.6217
P( 0  z  k ) = 0.1217
k = 0.31
Example 4
If x ~ N(24, 9) evaluate the following :
(a) P( 23.1  x  26.7 )
= P(
23.1  24
26.7  24
z
)
3
3
= P(  0.3  z  0.9 )
= 0.1179 + 0.3159
= 0.4338
(b) P(22.5<x<25.8)
=P(
22.5  24
25.8  24
z
)
3
3
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Chapter 16 The Normal Distribution and Its Applications
=P(  0.5  z  0.6 )
= 0.1915 + 0.2257
= 0.4172
(c) P( x  28.2 )
= P( z 
28.2  24
)
3
= P( z  1.4 )
= 0.5–0.4192
= 0.0808
(d) P( x  27.3 )
= P( z 
27.3  24
)
3
= P( z  1.1 )
= 0.5 + 0.3643
= 0.8643
Example 5
If x ~ N(34, 25), evaluate
(a) P( 28  x  32 )
= P(
28  34
32  34
z
)
5
5
= P(  1.2  z  0.4 )
= 0.3849 – 0.1554
= 0.2295
(b) P( x  37.1)
P( z 
37.1  34
)
5
= P( z  0.62 )
= 0.5 – 0.2324
= 0.2676
(c) P( x  26.8 )
P( z 
26.8  34
)
5
= P( z  1.44 )
= 0.5 – 0.4251
= 0.0749
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Chapter 16 The Normal Distribution and Its Applications
(d) P( x  30 | x  26.8)
30  34
26.8  34
|z
)
5
5
= P( z  0.8 | z  1.44 )
= P( z 
=P
=
0.4251  0.2881
0.5  0.4251
0.137
0.9251
= 0.1481 (to 4 d.p)
Example 6
If x ~ N(50, 100), find the values of k if
(a) P( 46  x  k ) = 0.6530
P(
46  50
k  50
z
) = 0.6530
10
10
P(  0.4  z 
k  50
) = 0.6530
10
k  50
= 2.82
10
k = 78.2
(b) P( k  x  52 ) = 0.2881
P(
k  50
52  50
z
) = 0.2881
10
10
P(
k  50
 z  0.2 ) = 0.2881
10
k  50
= –0.55
10
k = 44.5
P(
k  50
52  50
z
) = 0.2881
10
10
P(
k  50
 z  0 ) = 0.2088
10
k  50
= –0.55
10
k = 44.5
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F7 Mathematics and Statistics
Chapter 16 The Normal Distribution and Its Applications
Example 7
If x ~ N(12, 4), find the value of k if
P( 7  x  k ) = 0.8176
P( 7  x  k ) = 0.8176
P(
7  12
k  12
z
) = 0.8176
2
2
P(  2.5  z 
k  12
) = 0.8176
2
k  12
= 0.93
2
k = 13.86
Example 8
If z ~ N(0, 1). solve k for the following to 4 decimal places:
(a) P( 0  z  k ) = 0.4
k = 1.28 + 0.01(
0.4000  0.3997
)
0.4015  0.3997
k = 1.28 + 0.01(
3
)
18
= 1.2817 (to 4 d.p)
(b) P( 0  z  k ) = 0.25
P( 0  z  k ) = 0.67 + 0.01(
14
)
31
= 0.6745 (to 4 d.p)
(c) P( k  z  0 ) = 0.365
k = -1.10 + (
= -1.1032
7
)0.01
22
(to 4 d.p)
(d) P( k  z  0 ) = 0.4324
k = -1.49 + 0.01(
5
)
13
= -1.4938 (to 4 d.p)
(e) P( 0  z  k ) = 0.1057
k = 0.26 + 0.01(
31
)
38
= 0.2682 (to 4 d.p)
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Chapter 16 The Normal Distribution and Its Applications
§ 16.4
Problems involving normal distribution
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正態分佈應用題
Example 9
The lengths of sardine(沙甸魚) follow (依循) a normal distribution of mean 10cm and standard
deviation 1.2 cm. If a sardine can (罐) contains sardine with lenghts range (範圍) from 9.4cm to
10.9cm, find the probability that a randomly selected sardine are unsuitable (不適合) for can
food.
Let x be the length of sardine
x ~ N(10, 1.22)
P(suitable)
= P( 9.4  x  10.9 )
= P(
9.4  10
10.9  10
z
)
1.2
1.2
= P(  0.5  z  0.75 )
= 0.1915 + 0.2734
= 0.4649
 P(unsuitale) = 1 – 0.4649
= 0.5351
Example 10
The amount of soft drink (飲料) a machine discharges (排出) follow a normal distribution with
mean 8.oz. and standard deviation 0.2 oz. If the capacity (容量) of the cup is 8.28 oz.,
(a) find the probability that a particular cup of soft drink overflows (滿溢).
Let X of be amout of soft drink discharges per cup
x ~ N(8, 0.22)
P( x  8.28 )
=P(z 
8.28  8
)
0.2
= P( z  1.4 )
= 0.5 – 0.4192
= 0.0808
(b)
find the probability that 2 out of the 5 cups of soft drink discharged by the machine
overflows.
Reqired probability = C52(0.0808)2 (0.9192)3
= 0.0507 (to 4d.p)
(c)
find the probability that the 2nd overflow found on the 5th discharge.
Required probability = (0.0808)[C4(0.0808)(0.9192)3]
= 0.0203 (to 4 d.p)
(d)
find the adjusted (調節) capacity of the cup so that only 1% of the cups overflow.
k = adjusted capacity of the cup
P(x>k) = 0.01
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Chapter 16 The Normal Distribution and Its Applications
P( z 
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k 8
) =0.49
0.2
k 8
 2.3267
0.2
k = 8.4653 (to 4d.p)
Example 11
The amount of chemiacl (化學用品) used per week in a certain factory is a normal variable (正
態變量) with mean 1800 kg and standard deviation (標準變量) with mean 1800 kg and standard
deviation (標準差) 200kg. If the chemical used in the factory orders once a week (一週落一次
單), find the minium stock (存貨) the company should order at the start of a week to ensure that
there is at most a probability of 0.0017 of running out of stock (不夠存貨).
Let x be the amount of chemical used per week
x ~ N (1800, 2002)
k = order per week
P( x  k )  0.0017
P( 0  z 
k  1800
)  0.4983
200
k  1800
 2.93
200
k  2386
 The minium stock order is 2386 kg
Example 12
The liferime of a particular type of batteries (電池) follows a normal distribution with mean 1500
hours and a standard deviation of 150 hours.
(a) Find the interquartile range (四分位數間距) of the lifetime of the batteries.
(b) Find the proportion (比例) of batteries with lifetime at least (不少於) 1380
hours.
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Chapter 16 The Normal Distribution and Its Applications
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(c) Suppose a machine is powered (發動) by 3 of such batteries, and the machine
works (工作) if at least 2 of the three batteries with lifetimes at least 1380
hours. Find the probability that the machine works.
(d) Given that the machine works, find the probability that all the three batteries
with lifetimes at least 1380 hours.
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Chapter 16 The Normal Distribution and Its Applications
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Example 13
The number of customers served (服務) in a particular restaurant daily is a normal varibale (正態
變量) with mean 156 and standard deviation (標準差) of 18. Staffs receive a special payment (花
紅) if the number of customers in that day exceeds (超越) 185. Find the probabilites that
(a) staffs will receive a special payment in a particular day.
x: no. of customers served per day
x ~ N(156,182)
P( x  185.5 )
185.5  156
)
18
= 0.5  P(0  z  1.64)
= 0.5  P(0  z 
= 0.5 – 0.4495
= 0.0505
(b)
the number of customers served in a particular day lies between 149.5 and 164.5
inclusively (包括在內).
P (149.5  x  164.5)
149.5  156
164.5  156
z
)
= P(
18
18
= P(0.36  z  0.47)
= 0.1406 + 0.1808
= 0.3214
(c)
the fourth day of a particular week is the first day that staffs will receive a special
payment.
Required probability
= (0.9495)3 (0.0505)
= 0.0432 (to 4 d.p)
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Chapter 16 The Normal Distribution and Its Applications
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Example 14
The number of words per page in a particular book is a normal variable with mean
700 and standard deviation of 10 words. The book contains 250 pages.
(a)
Find the probability that the number of words in a particular page lies between 695 and
710 inclusively.
(b)
Find the probability that the fifth page(第五頁) is the second page with number of words
between 675 and 710 inclusively.
(c)
Find the expected number of pages with number of words between 695 and 710
inclusively.
(d)
Find the probability that exactly 2 of the first 5 pages with number of words between 695
and 710 inclusively.
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