1. “Teach A Level Maths” Statistics 2
Standardizing to z
© Christine Crisp

2. We can’t have a table of probabilities for every possible mean and variance ( as we would need an infinite number of tables! )
So, we always standardise to
( Mean = 0, variance = 1 ). This is easy to do.
If X is a random variable with distribution
then, if
Since this formula holds for X, it also holds for all the values of X, given by x.
The rule is
“subtract the mean and divide by the standard deviation”

3. Using the calculator

4. x = 400, so e.g.1 If X is a random variable with distribution
find (a) (b)
Solution: (a)
x = 400, so
Tables only give 2 d.p. for z so this is all we need.
So,

5. e.g.1 If X is a random variable with distribution
find (a) (b)
Solution: (b)
There are 2 values to convert so we use subscripts for z.
N.B. This is left of the mean so the z value will be negative.
So,

6. e.g.1 If X is a random variable with distribution
find (a) (b)
Solution: (b)

7. Tip: The diagrams for X and Z show the same areas so I don’t always draw both. If the question is straightforward I draw only the Z diagram but if I’m not sure what to do I’ll draw the X diagram ( and maybe the Z one as well ).
SUMMARY
To use tables to solve problems, we convert the values of the random variable X to values of the standardised normal variable using
We need to be careful not to confuse standard deviation and variance.
e.g means s = 4.

8. e. g. 2 A batch of batteries is claimed to last for 24 hours
e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?
Solution:
Let X be the random variable “ life of battery ( hours )”

9. e. g. 2 A batch of batteries is claimed to last for 24 hours
e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?
Solution:
Let X be the random variable “ life of battery ( hours )”
We want to find

10. e. g. 2 A batch of batteries is claimed to last for 24 hours
e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?
Solution:
Let X be the random variable “ life of battery ( hours )”
We want to find
So,
Approximately 20% do not last for 24 hours.

11. Exercise
1. If X is a random variable with distribution
find (a) (b) (c)
2. A shop sells curtain rails labelled 90 cm. In fact the lengths are normally distributed with mean 90·2 cm. and standard deviation 0·4 cm. What percentage of the rails are shorter than 90 cm ?

12. Solutions:
(a) 1.
( This is 1 standard deviation above the mean. )

13. Solutions: 1.
(b)

14. Solutions: 1.
(c)

15. 2. A shop sells curtain rails labelled 90 cm
2. A shop sells curtain rails labelled 90 cm. In fact the lengths are normally distributed with mean 90·2 cm. and standard deviation 0·4 cm. What percentage of the rails are shorter than 90 cm ?
Solution:
Let X be the random variable “length of rail (cm)”
We want to find
Approximately 31% are shorter than 90 cm.

17. 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.

18. We can’t have a table of probabilities for every possible mean and variance ( as we would need an infinite number of tables! )
So, we always standardise to ( Mean = 0, variance = 1 ). This is easy to do.
If X is a random variable with distribution
then, if
The rule is
“subtract the mean and divide by the standard deviation”
Since this formula holds for X, it also holds for all the values of X, given by x.

19. SUMMARY
We need to be careful not to confuse standard deviation and variance.
e.g means s = 4.
To use tables to solve problems, we convert the values of the random variable X to values of the standardised normal variable using

20. x = 400, so e.g.1 If X is a random variable with distribution
Tables only give 2 d.p. for z so this is all we need.
Solution: (a)
x = 400, so
e.g.1 If X is a random variable with distribution
find (a) (b)

21. Solution: (b)

22. e. g. 2 A batch of batteries is claimed to last for 24 hours
e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?
Solution:
Let X be the random variable “ life of battery ( hours )”
We want to find
So,
Approximately 20% do not last for 24 hours.