Transcript normal distribution

Normal Distribution
Normal distributions are a family of
distributions that have the same general shape.
•
•
•
•
•
They are symmetric with scores
more concentrated in the middle than
in the tails
Normal distributions are sometimes
described as bell shaped.
The area under each curve is the
same.
The height of a normal distribution
can be specified mathematically in
terms of two parameters: the mean
( ) and the standard deviation ( ).
The height (ordinate) of a normal
curve is defined as:
s
m
Equation
f ( x) =
1
s 2p
- ( x- m ) 2
e
2s
2
Features
•
•
•
•
•
It is bell-shaped
It is symmetrical about the mean
It extends from -∞ to +∞
The total area under the curve is 1
The maximum value of f(x) is
1
s 2p
Approximately 95% of the distribution lies within
two standard deviations from the mean.
Approximately 99.9% of the distribution lies
within three standard deviations from the mean.
The shape depends on the
value of s
Definition
The standard normal distribution is a
normal distribution with a mean of 0 and a
standard deviation of 1
Normal distributions can be transformed to standard
normal distributions by the formula:
z=
X -m
s
where X is a score from the original normal
distribution, m is the mean of the original normal
distribution, and s is the standard deviation of
original normal distribution.
The standard normal distribution is
sometimes called the z distribution. A z
score always reflects the number of
standard deviations above or below the
mean a particular score is.
For instance, if a person scored a 70 on a test with a
mean of 50 and a standard deviation of 10, then they
scored 2 standard deviations above the mean.
Converting the test scores to z scores, an X of 70
would be:
70 - 50
z=
=2
10
So, a z score of 2 means the original score was 2
standard deviations above the mean.
Note that the z distribution will only be a
normal distribution if the original
distribution (X) is normal.
Applying the formula will always produce a
transformed distribution with a mean of zero
and a standard deviation of one. However, the
shape of the distribution will not be affected
by the transformation.
Using the chart
• Need to know how
many standard
deviations you are
from the mean.
• Use
z=
x -m
s
Readings can be made
to the left ‘P’ (Chart 0.5 +)
P(Z <1.377) =
Readings can be made
to the left ‘P’ (Chart 0.5 +)
P(Z < 1.377) =
0.9158
To the centre ‘Q’
P(0 < Z <1.377) =
To the centre ‘Q’
P(0 < Z < 1.377) =
0.4158
Or to the right ‘R’ (Chart 0.5 - x)
P( X >1.377) =
Or to the right ‘R’ (Chart 0.5 - x)
P( X >1.377) =
0.0842
Lengths of metal strips produced by a machine
are normally distributed with a mean length of
150 cm and a standard deviation of 10 cm.
• Find the probability
that the length of a
randomly selected
strip is shorter than
165 cm.
165 -150
z=
= 1.5
10
P( X <165) = P(Z <1.5) = 0.9332
Lengths of metal strips produced by a machine
are normally distributed with a mean length of
150 cm and a standard deviation of 10 cm.
• Find the probability
that the length of a
randomly selected
strip is within 5 cm
of the mean
150
155 -150
z=
= 0.5
10
P(145 < X < 155) = P( Z < 0.5) = 2 ´ 0.1915
= 0.383
The time taken by the milkman to deliver to the High
Street is normally distributed with a mean of 12 mins
and standard deviation of 2 mins. He delivers milk every
day.
• Estimate the
number of days
during the year
when he takes
longer than 17 mins.
17 -12
z=
= 2.5
2
12
P( X >17) = P(Z > 2.5) = 0.5 - 0.4938 = 0.0062
Two days
The time taken by the milkman to deliver to the High
Street is normally distributed with a mean of 12 mins
and standard deviation of 2 mins. He delivers milk every
day.
• Estimate the
number of days
during the year
when he takes less
than ten mins.
10 -12
z=
= -1
2
12
P( X <10) = P(Z <1) = 0.5 - 0.3413 = 0.1587
58 days
The time taken by the milkman to deliver to the High
Street is normally distributed with a mean of 12 mins
and standard deviation of 2 mins. He delivers milk every
day.
• Estimate the
number of days
during the year
when he takes
between nine and
13 mins.
228 days
12
æ 9 -12
13 -12 ö
Pç
<z<
÷ = P(-1.5 < Z < 0.5)
2 ø
è 2
= 0.6247
Inverse Normal
The heights of female students at a particular
school are normally distributed with a mean of
169 cm and a standard deviation of 9 cm
• Given that 80% of
these female
students have a
height less than h
cm, find the value of
h.
• Given that 60% of
these female
students have a
height greater than s
cm, find the value of
s.
z=
x -m
s
• z = 0.842
h -169
0.842 =
9
169 h
h =169 + 9 ´ 0.842 =176.38
z=
x -m
s
• z = 0.253
s -169
-0.253 =
9
s 169
h =169 - 9 ´ 0.253 =166.723
Batteries for a transistor radio have a mean life
under normal usage of 160 hours, with a
standard deviation of 30 hours. Assuming a
normal distribution:
• Calculate the
percentage of
batteries which have
a life between 150
hours and 180
hours.
37.8%
Batteries for a transistor radio have a mean life
under normal usage of 160 hours, with a
standard deviation of 30 hours. Assuming a
normal distribution:
• Calculate the range,
symmetrical about
the mean, within
which 75% of the
battery lives lie.
125.5, 194.5
The masses of boxes of oranges are normally
distributed such that 30% of them are greater than 4.00
kg and 20% are greater than 4.53 kg. Estimate the
mean and standard deviation of the masses.
3.13, 1.67
The speeds of cars passing a certain point on a
motorway can be taken to be normally distributed.
Observations show that of cars passing the point, 95%
are travelling at less than85 kph and 10% are travelling
at less than 55 kph.
• Find the average
speed of the cars
passing the point.
68 kph
The speeds of cars passing a certain point on a
motorway can be taken to be normally distributed.
Observations show that of cars passing the point, 95%
are travelling at less than85 kph and 10% are travelling
at less than 55 kph.
• Find the proportion
of cars that travel at
more than 70 kph.
0.4282
Sometimes the normal distribution (a continuous
distribution) is used to approximate situations that are
really discrete. This occurs when data is measured to
the nearest whole number.
• The distribution
takes on the shape
of a normal
distribution. In fact,
the normal curve
was instigated by
De Moivre as an
approximation to the
Binomial.
The discrete data is represented by its
limits
• E.g. 7 becomes the
interval
6.5 < 7 < 7.5
Normal Approximation to the
Binomial Distribution
Notice that as N increases, the binomial
distribution approximate to a normal
distribution.
Binomial distributions
N = 5, p = 0.5
N = 5, p = 0.2
0.35
Binomial: N=5, p = 0.2
0.45
0.3
0.4
0.25
0.35
0.3
0.2
0.25
0.15
0.2
0.15
0.1
0.1
0.05
0.05
0
0
0
1
2
3
4
5
0
N = 10, p = 0.2
Binomial N=10, p = 0.2
1
2
3
4
5
N = 10, p = 0.5
Binomial: N= 10, p =0.5
0.35
0.3
0.3
0.25
0.25
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0
0
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
Binomial distributions
N = 20, p = 0.2
0.25
0.2
N = 20, p = 0.5
0.18
0.2
0.16
0.14
0.15
0.12
0.1
0.08
0.1
0.06
0.05
0.04
0.02
0
0
0
0.2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
N = 30, p = 0.2
3
0.16
0.18
4
5
6
7
8
9
10
11
12
13
14
15
16
17
N = 30, p = 0.5
0.14
0.16
0.12
0.14
0.1
0.12
0.1
0.08
0.08
0.06
0.06
0.04
0.04
0.02
0.02
0
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
The binomial distribution can be
approximated by a normal distribution under
the conditions
np and n(1- p ) are greater than 5
Careful of the language
• As a Binomial is a discrete distribution, a
continuity correction is necessary.
P( X £ 3) becomes P( X < 3.5)
• P(at most 3)
• P(fewer than 3) P( X < 3) becomes P( X < 2.5)
• P(exactly 3)
P( X = 3) becomes P(2.5 < X < 3.5)
• P(more than 3) P( X > 3) becomes P( X > 3.5)
• P(at least 3)
P( X ³ 3) becomes P( X > 2.5)
Example
• It is given that 40% of the population
support the Gambage Party. 150
members of the population are selected
at random. Use a suitable
approximation to find the probability that
more than 55 out of these 150 support
the Gambage Party.
It is given that 40% of the population support the
Gambage Party. 150 members of the population are
selected at random. Use a suitable approximation to
find the probability that more than 55 out of these 150
support the Gambage Party.
m = 60
•
•
•
•
•
Binomial distribution
N = 150, p = 0.4
Np = 60
Np(1-p) = 90
Use a normal
distribution
s = 150 ´ 0.4 ´ 0.6 = 6
55.5 - 60
z=
= -0.75
6
P( X > 55) = 0.7734
Normal approximation to the
Poisson distribution
Notice as values of l
increase, the distribution becomes normally
distributed.
l
l
l
l
=2
=3
=5
= 10
As lambda increase, the normal approximation
gets better.
We use the criteria
l >15
Poisson is a discrete
distribution and hence we
need to use a continuity
correction
The number of bacteria on a plate follows
a Poisson distribution with a parameter
60.
Find the probability
that there are
between 55 and 75
bacteria on a plate.
The number of bacteria on a plate follows
a Poisson distribution with a parameter
60.
Find the probability
that there are
between 55 and 75
bacteria on a plate.
l >15
The number of bacteria on a plate follows
a Poisson distribution with a parameter
60.
z1 =
z2 =
55.5 - 60
60
74.5 - 60
60
= -0.589
= 1.872
P(55 < X < 75) = 0.6887
The number of bacteria on a plate follows
a Poisson distribution with a parameter
60.
A plate is rejected if
less than 38
bacteria are found. If
2000 of such plates
are reviewed, how
many will be
rejected?
The number of bacteria on a plate follows
a Poisson distribution with a parameter
60.
z=
37.5 - 60
60
= -2.905
P( X < 38) = 0.00183
Number rejected = 4
Sums and differences of
normally distributed
random variables
When two random variables are
added, their sum is another
random variable.
E(T ) = E( X + Y ) = E( X ) + E(Y )
VAR(T ) = VAR( X + Y ) = VAR( X ) + VAR(Y )
Masses of a particular toy are normally
distributed with mean 20g and standard
deviation 2g. A random sample of 12 such
articles is chosen. Find the probability that the
total mass is greater than 230g.
• Each toy mass is
treated as an
independent value.
E(T ) = E( M1 + M 2 + … + M12 )
= 20 + 20 + … + 20 = 240g
Masses of a particular toy are normally
distributed with mean 20g and standard
deviation 2g. A random sample of 12 such
articles is chosen. Find the probability that the
total mass is greater than 230g.
• Each toy mass is
treated as an
independent value.
E(T ) = E( M1 + M 2 + … + M12 )
= 20 + 20 + … + 20 = 240g
VAR(T ) = VAR( M1 + M 2 + … M12 )
= 22 + 22 + … + 22 = 48
s (T ) = 48 = 6.928
Masses of a particular toy are normally
distributed with mean 20g and standard
deviation 2g. A random sample of 12 such
articles is chosen. Find the probability that the
total mass is greater than 230g.
z=
230 - 240
48
= -1.443
P(T > 230) = 0.9255
When two random variables are
subtracted, their sum is another
random variable.
E(T ) = E( X - Y ) = E( X ) - E(Y )
VAR(T ) = VAR( X - Y ) = VAR( X ) + VAR(Y )
A machine produces rubber balls whose diameters are
normally distributed with a mean of 5.50 cm and
standard deviation 0.08 cm. The balls are packed in
cylindrical tubes whose inside diameters are normally
distributed with mean 5.70 cm and standard deviation
0.12 cm.
If a randomly selected
ball is placed in a
randomly selected
tube, what is the
probability that the
clearance is
between 0.05 cm
and 0.25 cm.
A machine produces rubber balls whose diameters are
normally distributed with a mean of 5.50 cm and
standard deviation 0.08 cm. The balls are packed in
cylindrical tubes whose inside diameters are normally
distributed with mean 5.70 cm and standard deviation
0.12 cm.
E(C) = E(T ) - E(B) = 0.2
If a randomly selected
VAR(T - B) = 0.082 + 0.122 = 0.0208 ball is placed in a
randomly selected
tube, what is the
probability that the
clearance is
between 0.05 cm
and 0.25 cm.
A machine produces rubber balls whose diameters are
normally distributed with a mean of 5.50 cm and
standard deviation 0.08 cm. The balls are packed in
cylindrical tubes whose inside diameters are normally
distributed with mean 5.70 cm and standard deviation
0.12 cm.
E(C) = E(T ) - E(B) = 0.2
If a randomly selected
VAR(T - B) = 0.082 + 0.122 = 0.0208 ball is placed in a
0.05 - 0.2
randomly selected
z1 =
= -1.040
0.0208
tube, what is the
0.25 - 0.2
probability that the
z2 =
= 0.347
0.0208
clearance is
between 0.05 cm
and 0.25 cm.
A machine produces rubber balls whose diameters are
normally distributed with a mean of 5.50 cm and
standard deviation 0.08 cm. The balls are packed in
cylindrical tubes whose inside diameters are normally
distributed with mean 5.70 cm and standard deviation
0.12 cm.
E(C) = E(T ) - E(B) = 0.2
If a randomly selected
VAR(T - B) = 0.082 + 0.122 = 0.0208 ball is placed in a
0.05 - 0.2
randomly selected
z1 =
= -1.040
0.0208
tube, what is the
0.25 - 0.2
probability that the
z2 =
= 0.347
0.0208
clearance is
between 0.05 cm
P(0.05 < C < 0.25) = 0.4865
and 0.25 cm.
Multiples of Independent Normal
Variables.
E(aX + bY ) = aE( X ) + bE(Y )
VAR(aX + bY ) = a VAR( X ) + b VAR(Y )
2
2
Great care must be taken in
distinguishing between a sum
of random variables and a
multiple of a random variable.
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
• A bottle of each size
is selected at
random. Find the
probability that the
large bottle contains
less than four times
the amount in the
small bottle.
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
--
´4
• A bottle of each size
is selected at
random. Find the
probability that the
large bottle contains
less than four times
the amount in the
small bottle.
P(L - 4S < 0)
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
--
´4
E(L - 4S ) = E(L) - 4E(S )
= 1012 - 4 ´ 252 = 4
P(L - 4S) < 0
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
E(L - 4S ) = E(L) - 4E(S )
= 1012 - 4 ´ 252 = 4
VAR(L - 4S ) = VAR(L) + 42 VAR(S )
= 25 + 16 ´ 4 = 89
--
´4
z=
0-4
= -0.424
89
P(L - 4S < 0) = 0.3358
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
• One large bottle and
four small bottles
are selected at
random. Find the
probability that the
amount in the large
bottle is less than
the total amount in
the four small
bottles.
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
• One large bottle and
four small bottles
are selected at
random. Find the
probability that the
amount in the large
bottle is less than
the total amount in
the four small
bottles.
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
P(L < S1 + S2 + S3 + S4 )
= P(L - (S1 + S2 + S3 + S4 ) < 0)
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
E(L - (S1 + S2 + S3 + S4 ) =
E(L) - E(S1 + S2 + S3 + S4 ) =
1012 -1008 = 4
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
E(L - (S1 + S2 + S3 + S4 ) =
E(L) - E(S1 + S2 + S3 + S4 ) =
1012 -1008 = 4
VAR(L - (S1 + S2 + S3 + S4 ) =
VAR(L) + VAR(S1 + S2 + S3 + S4 ) =
25 + 4 + 4 + 4 + 4 = 41
A soft drinks manufacturer sells bottles of
drinks in two sizes. The amount in each
bottle is normally distributed.
Mean (ml) Variance
Small
252
4
Large
1012
25
z=
0-4
= -0.625
41
P(L - (S1 + … + S4 ) < 0) =
0.2661
Check that you know the
difference in these two
questions