chapter 04 - normal distribution

8/13/2019 Chapter 04 - Normal Distribution

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S t a t i s t i c s f o r B u s i n e s s

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C h a p t e r 0 4 : T h e N o r m a l D i

s t r i b u t i o n

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STATISTICS FOR BUSINESS [IUBA]

CHAPTER 04

THE NORMAL DISTRIBUTION

Standard Normal Dist ribut ion Normal Dist ribut ion

~(,) ~(,)

PART I

Finding probabilities of the normal distribution with given values

Step 01: Use the following formula to transform the normal random variable

,

where ~(

,) into the standard normal random variable , where~(0,1).

= −

Step 02: Use the calculator or Table 2 (Areas of Standard Normal Distribution) in

Appendix C to compute the probabilities (or areas) of the normal

distribution based on the standard normal distribution.






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C h a p t e r 0 4 : T h e N o r m a l D i s t r i b u t i o n

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CALCULATOR INSTRUCTION

Step No.01: Press MODE 3: STAT [ AC ]

Step No.02: Press SHIFT + 1 | [ STAT ] 7: DISTR

After that, the calculator will show you 4 available symbols.

However, we just pay attention to the first three ones.

1 : P ( 2 : Q ( 3 : R (

+ We use [ 1 : P ( ] to compute the probability between the

standard normal random variable to −∞, or ( < )

+ We use [ 2 : Q ( ] to compute the probability between the

standard normal random variable to the mean or (0 < < ).

+ We use [ 3 : R ( ] to compute the probability between the

standard normal random variable to +∞. ( > )

Step No.03: Press [ = ] to get the result.






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Example | Part 1:

(Case of f inding probabilit ies of the normal dist ribut ion with given values)

PROBLEM:

A psychologist has devised a stress test for dental patients sitting in the waiting rooms.

According to this test, the stress scores (on a scale of 1 to 10) for patients waiting for root canal

treatments are found to be approximately normally distributed with a mean of 7.59 and a

standard deviation of 0.73.

a. What percentage of such patients have a stress score lower than 6.0?

SOLUTION: = 7.59, = 0.73

Step 01: ( < 6.0) = <... = ( < −2.1781)

Step 02: MODE 3 : STAT [ AC ]

SHIFT + 1| [STAT] 7 : DISTR 1 : P (-2.1781)

Then, press [ = ] to get the result (−.) = .

Summing up, ( < 6.0) = ( < −2.1781) = 0.0147

b. What is the probability that a randomly selected root canal patient sitting in the waiting

room has a stress score between 7.0 and 8.0?

SOLUTION: = 7.59, = 0.73

Step 01: (7.0 < < 8.0) = ... < <

...

=

−0.8082<

<0.5616)

Step 02: (There are three different methods of using the pocket calculator

to solve the problem of calculating the probabilit y between two

given values. You can use only one of three following methods

that depends on your choice, not al l of t hem, since each of them

always provides the same result wi th others)






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METHODS 01: Using 1 : P (

MODE 3 : STAT [ AC ]

SHIFT + 1| [STAT]

7 : DISTR

1 : P (0.5616)

[ ]

SHIFT + 1| [STAT] 7 : DISTR 1 : P (-0.8082)

Then, press [ = ] to get the result

(.) − (−.) = .

METHODS 02: Using 3 : R (


SHIFT + 1| [STAT] 7 : DISTR 3 : R (-0.8082)

[ ]

SHIFT + 1| [STAT] 7 : DISTR 3 : R (0.5616)


(−.) − (.) = .

METHODS 03: Using 2 : Q (


SHIFT + 1| [STAT] 7 : DISTR 2 : Q (-0.8082)

[ + ]

SHIFT + 1| [STAT] 7 : DISTR 2 : Q (0.5616)


(−.) +(.) = .






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Summing up,

(7.0 < < 8.0) = (−0.8082< < 0.5616) = 0.5033

c. The psychologist suggests that any patient with a stress score of 9.0 or higher should be given

a sedative prior to treatment. What percentage of patients waiting for root canal treatments

would need a sedative if this suggestion is accepted?

SOLUTION: = 7.59, = 0.73

Step 01: ( ≥ 9.0) = ≥ (... = ( ≥ 1.9315)

Step 02: MODE 3 : STAT [ AC ]

SHIFT + 1| [STAT] 7 : DISTR 3 : R (1.9315)

Then, press [ = ] to get the result (.) = .

Summing up, ( > 9.0) = ( > 1.9315) = 0.0267






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PART II

Finding the values of X given a probability

Step 01: The probability that a normal random variable will be above (below, or

symmetric) its mean a certain number of standard deviations is exactly

equal to the probability that the standard normal random variable will be

above (below, or symmetric) its mean the same number of (its) standard

deviations.

In particular, ( > ) = ( > )

(

<

) =

(

<

)

( < < ) = (− < < )

Step 02: Find TA (Table Area)

(Table Area TA for a point of the standard normal distribution is the area

given in the standard normal probability table of Table 2 in Appendix C

under the standard normal curve between 0 and point

>0)

In particular,

For ( > ):

If ( > ) < ., =. − ( > )

If ( > ) > ., =( > ) − .

For ( < ):

If (

< ) < ., =. −

(

< )

If ( < ) > ., =( < ) − .

For (− < < ), = (− < <)/






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Step 03: Look inside Table 2 (Areas of the Standard Normal Distribution) in

Appendix E for the values of corresponding to TA.

Step 04: Use the following formula to transform the standard normal random

variable to the normal random variable = ±

In particular,

For ( > ):

If (

> ) < ., the value of “” will be positive, or +

If ( > ) > ., the value of “” will be negative, or −

For ( < ):

If ( < ) < ., the value of “” will be negative, or −

If ( < ) > ., the value of “” will be positive, or +

For

(

− <

<

), the values of “

” will include negative and positive

values, or – and +






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Example | Part 2:

(Finding t he values of X given a probabilit y )

PROBLEM 01:

If X is a normally distributed random variable with mean 120 and standard deviation 44, find a

value x such that the probability that X will be less than x is 0.56.

SOLUTION: We have ~(120,44)

Step 01: We are looking for the value of the random variable

such that

(

<

) =

.

. In order to find it, we look for the value of the

standard normal deviation such that ( < ) = ..

Step 02: If the area to the left of is equal to 0.56, the area between 0 and (the

Table Area) is equal to

= . − . = ..

Step 03: We look inside Table 2 (Areas of Standard Normal Distribution) in

Appendix C for the value corresponding to = 0.06 and find = 0.15 (actually, = 0.0596, which is close enough to 0.06).

Step 04: We need to find the appropriate value. Here we use the following

equation.

=

+

=

+(

.

)(

) =

.






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PROBLEM 02:

For a normal random variable with mean 19,500 and standard deviation 400, find a point of the

distribution such that the probability that the random variable will exceed this value is 0.02.

SOLUTION: =19,500, = 400

( > ) = ( > ) = 0.02

= 0.5 − 0.02 = 0.48

= 2.05

Thus, = + = 19,500+(2.05)(400) = 20,320

PROBLEM 03

For ~ (32,7) , find two values and , symmetrically lying on each side of the mean,

with ( < < ) = 0.99.

SOLUTION: = 32, =7

( < < ) = (− < < ) = 0.99

=0.99/ 2 = 0.495

= 2.576

Thus, = +(−) = 32+ (−2.576)(7) = 13.968

= +() = 32+ (2.576)(7) = 50.032






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PART III

The Normal Approximation to the Binomial Distribution

Condition:

Normal Dist ribut ion as an Approximat ion to Binomial Dist ribut ion

Usually, the normal distribution is used as an approximation to the binomial distribution

when and (1 − ) are both greater than 5—that is, when

>

and

(

− ) >

Step 1: Compute and for the binomial distribution.

= = ( − )

Step 2: Convert the discrete random variable into a continuous random variable.

Continui ty Correction Factor

The addition of .5 and/or subtraction of .5 from the value(s) of when the normal

distribution is used as an approximation to the binomial distribution, where is the

number of successes in trials, is called the continuity correction factor.

Step 3: Compute the required probability using the normal distribution.






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Example | Part 3:

(The Normal Approximat ion to t he Binomial Distribut ion )

PROBLEM:

CASIO Vietnam makes calculators. Consumer satisfaction is one of the top priorities of the

company’s management. The company guarantees the refund of money or a replacement for

any calculator that malfunctions within two years from the date of purchase. It is known from

past data that despite all efforts, 5% of the calculators manufactured by this company

malfunction within a 2-year period. The company recently mailed 500 such calculators to its

customers.

a. Find the probability that exactly 29 of the 500 calculators will be returned for refund or

replacement within a 2-year period.

SOLUTION:

Step 1: Compute and for the binomial distr ibut ion.

= = 500×0.05 = 25

=

(1

− ) =

√ 500×0.05×0.95 =

√ 95

2 ≈ 4.8734

Step 2: Convert the discrete random variable into a cont inuous random variable.

To make the continuity correction, we subtract 0.5 from 29 and add 0.5 to 29, which gives the

interval 28.5 to 29.5 to obtain the value 29. Thus, ( = 29) for the binomial problem will be

approximately equal to (28.5 ≤ ≤ 29.5) for the normal distribution.

Step 3: Compute t he required probability using t he normal distr ibuti on.

(28.5 ≤ ≤ 29.5) = 28.5− 25√ 952 ≤ − ≤ 29.5 − 25√ 95

2

7√ 95

95 ≤ ≤ 9√ 95

95 =(0.7182 ≤ ≤ 0.9234) = 0.0584






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Thus, based on the normal approximation, the probability that exactly 29 of the 500 calculators

will be returned for refund or replacement within a 2-year period is approximately 0.0584

b. What is the probability that 27 or more of the 500 calculators will be returned for refund or

replacement within a 2-year period?

SOLUTION:


= = 500×0.05 = 25

=

(1

− ) =

√ 500×0.05×0.95 =

√ 95

2 ≈ 4.8734


For the continuity correction, we subtract 0.5 from 27, which gives 26.5 to obtain the value 27.

Thus, ( ≥ 27) for the binomial problem will be approximately equal to ( ≥ 26.5) for the

normal distribution.


( ≥ 26.5) = − ≥ 26.5 − 25√ 952

≥ 3√ 95

95 = ( ≥ 0.3078) = 0.3791

Thus, based on the normal approximation, the probability that 27 or more of the 500

calculators will be returned for refund or replacement within a 2-year period is approximately

0.3791






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c. What is the probability that 15 to 22 of the 500 calculators will be returned for refund or

replacement within a 2-year period?

SOLUTION:


= = 500×0.05 = 25

= (1 − ) = √ 500×0.05×0.95 =√ 95

2 ≈ 4.8734


For the continuity correction, we subtract 0.5 from 15 and add 0.5 to 22, which gives the

interval 14.5 to 22.5 to obtain the interval 15 to 22. Thus, (15 ≤ ≤ 22) for the binomialproblem will be approximately equal to (14.5 ≤ ≤ 22.5) for the normal distribution.


(14.5 ≤ ≤ 22.5) = 14.5− 25√ 952

≤ − ≤ 22.5 − 25√ 952

−21

√ 95

95 ≤ ≤ − √ 95

91 =

(

−2.1546

≤ ≤ −0.5130) = 0.2884

Thus, based on the normal approximation, the probability that 15 to 22 of the 500 calculators

will be returned for refund or replacement within a 2-year period is approximately 0.2884

chapter 04 - normal distribution

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