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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.
Reversing Normal Calculations
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 .
Reversing Normal Calculations
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
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.
10)( xXPWe want .
Reversing Normal Calculations
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
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.
z
Z 10
0
28161 z90 p
Reversing Normal Calculations
)( 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
Reversing Normal Calculations
Z
z 0
)( XP
Solution: Let X be the random variable “lifetime of battery ( hours)”
)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
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?
64491 So, z = -Tip: Write z = - as soon as you spot that z is negative so you don’t forget the minus sign.
Reversing Normal Calculations
)( XP
Solution: Let X be the random variable “lifetime of battery ( hours)”
)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)
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?
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.
Reversing Normal Calculations
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
)
Reversing Normal Calculations
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
Reversing Normal Calculations
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
Reversing Normal Calculations
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
Reversing Normal Calculations
Z
0
050
z
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:
Let X be the random variable “exam mark”
)20,50(~ 2NX
050
050)(050)( zZPxXP(b
)
Reversing Normal Calculations
Z
0
050
z
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:
050)(050)( zZPxXP64491 z
Let X be the random variable “exam mark”
)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
Reversing Normal Calculations
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
Reversing Normal Calculations
Reversing Normal Calculations
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.
Reversing Normal Calculations
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(~ 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
Reversing Normal Calculations
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,
Reversing Normal Calculations
Z
z 0
)( 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 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
Reversing Normal Calculations
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
Reversing Normal Calculations
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.