Reversing Normal Calculations Statistics 2. Reversing Normal Calculations In the previous...

Reversing Normal Calculations

Statistics 2

Reversing Normal CalculationsIn the previous presentation there was an

example where a claim was made that batteries would last 24 hours. However, 20% of them failed in under this time.This manufacturer would get a lot of complaints!However, in many industrial processes times or lengths or weights cannot be exact and some items will fall outside acceptable levels.

Suppose the battery manufacturer is prepared to accept that 5% will not last as long as is claimed and so wants to know what length of time should be claimed on the package.

We need to reverse the process we used before.

As statisticians we are being given the percentage and want to find the corresponding value of x.


Solution:

e.g.1 A random variable X has a Normal Distribution with mean 100 and standard deviation 15. Find the value of X that is exceeded by 10% of the distribution.

)15,100(~ 2NX

We now need to find z.

This is the same as

10)( zZP

Instead of having to use the table backwards there is a separate table written the other way round. It doesn’t give as many values but contains those you are likely to need. Find the table now. It’s called “Percentage Points of the Normal Distribution”

z

Z 10 p

0

10)( xXPWe want .


Solution: )15,100(~ 2NX

The table shows the area to the left of z .

z

Z90 p

90 pSo, we want

z 28161

This is the same as

10)( zZP

z

Z 10 p

0


10)( xXPWe want .


Solution: )15,100(~ 2NX

Now we need x:

xz

15

10028161

x

10)( xXPWe want .This is the same as

10)( zZP

Rearranging: 1001528161 x

10022419 x )..3(119 fsx


z

Z 10

0

28161 z90 p


)( XP

Solution: Let X be the random variable “lifetime of battery ( hours)”

e.g.2 The running time of a batch of batteries has a normal distribution with a mean time of 29 hours and standard deviation 6 hours. The manufacturer can accept that 5% of the batteries will fail to last the time he is going to claim on the packaging. What time should this be?

)6,29(~ 2NXWe want 050 x


Z

z 0

)( XP


)6,29(~ 2NXWe want

or 050)( zZP050 x

050 p

We just use 0·95 BUT the z we want is negative.

The smallest value of p in the tables is 0·5.

Can you see what to do?

950 050


64491 So, z = -Tip: Write z = - as soon as you spot that z is negative so you don’t forget the minus sign.


)( XP


)6,29(~ 2NXWe want

or 050)( zZP050 x

x

z6

2964491

x

29664491 xx 8694929

The time claimed should be 19 hours.

64491 z (negative)


119 x)..3( fs

Z

z 0

050 p

Reversing Normal CalculationsSUMMARY

To solve problems where we are given a probability, percentage or proportion we need to find values of the random variable.

If given a percentage or proportion we convert to a probability and find the z value using the table called “Percentage Points of the Normal Distribution”. Always draw a diagram and watch out for values of z to the left of the mean as they will be negative.

Use the standardizing formula to convert to x.

Z

0z

z is negative

e.g.

Write z = - immediately.


Exercise

80)( zZP

1. Find the values of Z corresponding to the following:

(a) (b) 010)( zZP

30)( zZP(c)

Solution:

Z

0 z

80 p

(a) 80)( zZP

84160 z

70)( zZP(d

)


Z

z

30

0

30

Solution:

(b)

Z

0 z

010

010)( zZP

32632 z

Use p = 0·99,

30)( zZP(c)

z is negative

z


Z

z

30

0

70 p

Solution:

(b)

Z

0 z

010

010)( zZP

32632 z

Use p = 0·99,

z

Use p = 0·7,

30)( zZP(c)

z is negative30

52440

Reversing Normal CalculationsSolutio

n:

z

z is negative

70)( zZP(d

)

Z

0z

70

Z

0

70

Use p = 0·7,

52440


Z

0 z

10

Exercise2. The marks of candidates in an exam are

normally distributed with mean 50 and standard deviation 20. (a) If 10% of candidates are to be given an A* what mark does this correspond to?(b) If 5% fail the exam, what is the pass mark?

Solution:

28161 z

Let X be the random variable “exam mark”

)20,50(~ 2NX

20

5028161

x

)..3(675 fsx ( With a mark of 75, just over 10% would get

the A* and with a mark of 76 slightly fewer. )

10)(10)( zZPxXP(a

)

90 p


Z

0

050

z



Solution:


)20,50(~ 2NX

050

050)(050)( zZPxXP(b

)


Z

0

050

z



Solution:

050)(050)( zZPxXP64491 z


)20,50(~ 2NX

950 p

20

5064491

x

)..3(117 fsx The pass mark would be

17.

(b)

050

Reversing Normal Calculationse.g.3 The mean weight of potatoes in a batch is 150

g and the standard deviation is 40 g. The potatoes are graded to be put into bags. The lightest 10% are discarded. The heaviest 20% are sorted to be sold separately. Between what range of weights are the potatoes in the bags?

10)( 1 zZP

Solution:Let X be the random variable “weight of a

potato (g)”)40,150(~ 2NX

20)( 2 zZP

We have 2 percentages which we deal with separately.

Z

01z

10 Z

0 2z

20


90 pZ

01z

10)( 1 zZP

)40,150(~ 2NX

20)( 2 zZP

Z

0 2z

2080 p

281611 z 841602 z

x

z

40

15028161 1

x

40

15084160 2

x

1504028161 1 x 1504084160 2 x

736981 x 6641832 x

The weights range from 99 g. to 184 g.

10


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


Solution:


)15,100(~ 2NX10)( xXPWe want .

We now need to find z.

This is the same as

10)( zZP

Instead of having to use the table backwards there is a separate table written the other way round. It doesn’t give as many values but contains those you are likely to need. Find the table now. It’s called “Percentage Points of the Normal Distribution”

z

Z 10

0


As before, the table shows the area to the left of z .

90 pSo, we wantz

Z 10

Now we need x:

xz

15

10028161

x

28161 z

Rearranging: 1001528161 x

10022419 x)..3(119 fsx

0

90

From the table,


Z

z 0

)( XP


e.g.2 The running time of a batch of batteries has a normal distribution with a mean time of 29 hours and standard deviation 6 hours. The manufacturer can accept that 5% of the batteries will not last the time written on the packaging. What time should this be?

)6,29(~ 2NXWe want

or 050)( zZP050 x

050

The table doesn’t list values below 0·5 so we need to find the value of z corresponding to p = 0·95 and change the sign.

64491 So, z =

050

Z

z 0

950 p050


x

z

6

2964491

x

29664491 x

x 8694929

The time claimed should be 19 hours.

119 x

Reversing Normal Calculationse.g.3 The mean weight of potatoes in a batch is 150

g and the standard deviation is 40 g. The potatoes are graded to be put into bags. The lightest 10% are discarded. The heaviest 20% are sorted to be sold separately. Between what range of weights are the potatoes in the bags?

10)( 1 zZP

Solution:Let X be the random variable “weight of a

potato (g)”)40,150(~ 2NX

20)( 2 zZP

We have 2 percentages which we deal with separately.

Z

01z

10 Z

0 2z

20


Z

01z

10

10)( 1 zZP

)40,150(~ 2NX

20)( 2 zZP

Z

0 2z

2080 p

281611 z 841602 z

x

z

40

15028161 1

x

40

15084160 2

x

1504028161 1 x 1504084160 2 x736981 x 6641832 x

The weights range from 99 g. to 184 g.

90 p

Reversing Normal CalculationsSUMMARY

To solve problems where we are given a probability, percentage or proportion we need to find values of the random variable. If given a percentage or proportion we convert to a probability and find the z value using the table called “Percentage Points of the Normal Distribution”. Always draw a diagram and watch out for values of z to the left of the mean as they will be negative.

Use the standardizing formula to covert to x.

Z

0z

e.g.

z is negativeWrite z = -

immediately.

Reversing Normal Calculations Statistics 2. Reversing Normal Calculations In the previous...

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