Addmath form 5 normal distribution not mine

1. NORMAL
DISTRIBUTION
****************
PROBABILITY DISTRIBUTIONS

2. Continuous random
variable:
Variable which can be
measured and take on any
value in a given range

3. The notation for
normal distribution is
)
,
( 2


N
iance
and
mean
where
var
2
=
=



4. Properties of normal distribution
• The graph is bell shaped and
symmetrical about the vertical line
passing through the mean,
• The total area under the graph is
equal to 1
• The graph extends indefinitely to the
right and the left of the mean,
but never touches or intersects the
x-axis


x
f(x)


5. The notation for a
standard normal
distribution is
)
1
,
0
(
~ N
Z
1
,
tan
0
,
=
=


deviation
dard
s
and
mean
with

6. Properties of standard normal
distribution
• The graph is bell shaped and
symmetrical about the vertical axis
where mean,
• The total area under the graph is
equal to 1
• The graph extends indefinitely to the
right and the left of the mean,
but never touches or intersects the
horizontal axis
0
=

0
=


z
f(z)
=0

7. Z is a random variable having the
standard normal distribution.
Evaluate each of the following:
)
86
.
1
(
)
)
684
.
2
(
)
)
0
057
.
1
(
)
)
527
.
0
(
)
)
128
.
2
425
.
1
(
)
)
93
.
0
(
)
)
693
.
1
605
.
0
(
)
)
2
.
0
(
)

−



−
−






−

Z
P
h
Z
P
d
Z
P
g
Z
P
c
Z
P
f
Z
P
b
Z
P
e
Z
P
a
Example 1

8. Solution 1
4207
.
0
)
2
.
0
(
)
=

Z
P
a
936
.
0
)
93
.
0
(
)
=

Z
P
b
2991
.
0
)
527
.
0
(
)
527
.
0
(
)
=

=
−

Z
P
Z
P
c
99682
.
0
)
684
.
2
(
1
)
684
.
2
(
)
=

−
=
−

Z
P
Z
P
d

9. Solution 1
6821
.
0
)
0452
.
0
727
.
0
(
1
)
693
.
1
605
.
0
(
)
=
+
−
=


− Z
P
e
0604
.
0
)
128
.
2
(
)
425
.
1
(
)
128
.
2
426
.
1
(
)
=

−

=


Z
P
Z
P
Z
P
f
3547
.
0
)
5
.
0
1453
.
0
(
1
)
0
057
.
1
(
)
=
+
−
=


− Z
P
g
0628
.
0
)
0314
.
0
(
2
)
86
.
1
(
)
=
=

Z
P
h

10. A random variable, Z , has the
standard normal distribution. Find
the value of k if
148
.
0
)
6
.
0
(
)
7
.
0
)
(
)
9247
.
0
)
(
)
0202
.
0
)
(
)
3974
.
0
)
(
)
=


=

=

=

=

k
Z
P
e
k
Z
P
d
k
Z
P
c
k
Z
P
b
k
Z
P
a
Example 2

11. Solution 2
26
.
0
3974
.
0
)
(
)
=

=

k
positive
be
must
k
k
Z
P
a
05
.
2
0202
.
0
)
(
)
−
=

=

k
negative
be
must
k
k
Z
P
b
438
.
1
0753
.
0
)
(
9247
.
0
)
(
1
9247
.
0
)
(
)
−
=

=

=

−
=

k
k
Z
P
k
Z
P
negative
be
must
k
k
Z
P
c

12. Solution 2
144
.
1
1263
.
0
)
(
148
.
0
)
(
2743
.
0
148
.
0
)
(
)
6
.
0
(
148
.
0
)
6
.
0
(
)
=

=

=

−
=

−

=


k
k
Z
P
k
Z
P
k
Z
P
Z
P
k
Z
P
e
524
.
0
3
.
0
)
(
7
.
0
)
(
1
7
.
0
)
(
)
=

=

=

−
=

k
k
Z
P
k
Z
P
positive
be
must
k
k
Z
P
d

13. A normal random variable,X with
and can be converted into
standard normal random variable,Z
with and by using the
formula:


−
=
X
Z
0
=


1
=


on
distributi
normal
a
of
deviation
dard
s
on
distributi
normal
of
mean
iable
random
normal
a
of
value
X
value
z
Z
tan
var
=
=
=
−
=



14. z-value are also
known as
standard score
or z-score

15. X is a random variable having a
normal distribution with a mean of
50 and standard deviation of 8.
Convert each of the following
x-values to a z-value:
a) x = 48
b) x = 56
Example 3

16. Solution 3
25
.
0
8
50
48
)
−
=
−
=
−
=


X
Z
a
75
.
0
8
50
56
)
=
−
=
−
=


X
Z
b

17. The masses of students of a school have
a normal distribution with mean 52 kg
and standard deviation 10 kg. A students
is selected randomly from the school.
Find the probability that the mass of
the selected student is
a) More than 50 kg,
b) Between 47 kg and 53.8 kg
Example 4

18. Solution 4
5793
.
0
4207
.
0
1
)
2
.
0
(
1
)
2
.
0
(
)
10
52
50
(
)
50
(
)
=
−
=

−
=
−

=
−

=

Z
P
Z
P
Z
P
X
P
a
10
52 =
= 
 and
student
a
of
masses
the
be
X
Let
2629
.
0
4286
.
0
3085
.
0
1
)
18
.
0
(
)
5
.
0
(
1
)
18
.
0
5
.
0
(
)
10
52
8
.
53
10
52
47
(
)
8
.
53
47
(
)
=
−
−
=

−

−
=


−
=
−


−
=


Z
P
Z
P
Z
P
Z
P
X
P
b

19. The lifespan of an electric bulb is
normally distributed with mean
100 hours and standard deviation
12 hours.
a) A bulb is chosen at random,
find the probability that its
lifespan is shorter than 90
hours.
b) If 95% of the bulbs lasted
greater than k hours, find the
value of k.
Example 5

20. Solution 5
2025
.
0
)
833
.
0
(
)
833
.
0
(
)
12
100
90
(
)
90
(
)
=

=
−

=
−

=

Z
P
Z
P
Z
P
X
P
a
12
100 =
= 
 and
bulb
electric
an
of
lifespan
the
be
X
Let
26
.
80
645
.
1
12
100
05
.
0
)
12
100
(
95
.
0
)
12
100
(
1
)
12
100
(
95
.
0
)
12
100
(
95
.
0
)
(
)
=
−
=
−

=
−

=
−

−
−
=
−

=

k
k
k
Z
P
k
Z
P
negative
be
must
k
k
Z
P
k
X
P
b

21. It is found that the length of one teaching
period in a school follows a normal
distribution with a mean of 35.5 minutes
and a standard deviation of 1.25 minutes.
a) Find the probability that for a teaching
period chosen at random, the teaching
time is more than 37 minutes.
b) If 78% of the teaching periods exceed m
minutes, find the value of m.
[5 marks]
PAST YEAR-2009

22. The volume of mineral water in bottles
produced by a factory follows a normal
distribution with a mean of 755 ml and a
standard deviation of 10 ml.
a) If a bottle of the mineral water is chosen at
random, find the probability that the volume
of mineral water in the bottle is less than
765 ml.
b) Given that 80% of the bottles has a volume
of mineral water more than k ml, find the
value of k.
[5 marks]
PAST YEAR-2012