Statistics 1 Revision Notes - Mr Barton Maths

We need the value of Z such that Φ(Z) = 0⋅01
From the tables Z = –2⋅3263 to 4 D.P. from tables (remember to look in the small table
after the Normal tables)
Standardising the variable
⇒ Z =
X − μ
σ
=
t
− 90
15
t − 90
⇒ = − 2 ⋅3263
to 4 D.P. from tables
15
⇒
t = 55⋅1 to 3 S.F.
0⋅01
z 0
so the manufacturer should give a guarantee period of 55 months (4 years 7 months)
Example: The results of an examination were Normally distributed. 10% of the candidates
had more than 70 marks and 20% had fewer than 35 marks.
Find the mean and standard deviation of the marks.
Solution:
First we need the values from the tables
⇒ Φ(–0⋅8416) = 0⋅2,
and 1 – Φ(1⋅2816) = 0⋅1
0⋅2
0⋅1
x
−0⋅8416
0
1⋅2816
Using Z
− μ
= X we have
σ
– 0⋅8416
= 35 − μ
σ
⇒ μ = 35 + 0⋅8416σ
− μ
and 1⋅2816 = 70
σ
⇒ μ = 70 – 1⋅2816σ
0⋅2
35 μ 70
0⋅1
x
⇒ σ = 16⋅5 and μ = 48⋅9 to 3 S.F.
simultaneous equations
40 14/04/2013 Statistics 1 SDB