Transcript Application of Standard Normal Distribution

Application of Standard Normal
Distribution
• Q1: A certain type of storage battery
lasts, on average 3.0 years with
standard deviation of 0.5 year,
assuming that battery life is normally
distributed; find the probability that a
given battery will last less than 2.3
years.
Application of Standard Normal
Distribution
•
Application of Standard Normal
Distribution
• Solution:
• We want to find P( X < 2.3) to find this
area or probability by computing z as
z
X 

2.3  3

 1.4
0.5
P(X < 2.3) = P( z < -1.4 )
From the normal distribution table we can
conclude that the probability is 0.0808
Application of Standard Normal
Distribution
• Q2: An electrical manufactures light
bulbs that have a life before burn out
that is normally distributed with mean
equal to 800 hours and a standard
deviation of 40 hours. Find the
probability that a bulb burns between
778 and 834 hours.
Application of Standard Normal
Distribution
Solution:
We should calculate z1 and z2 as:
z1 
778  800
834  800
 0.55 and z 2 
 0.85
40
40
Application of Standard Normal
Distribution
• Then
• P (778 < X< 834) = P (z < 0.85) – P (z < -0.55)
From table
P (z < 0.85) – P (z < -0.55) = 0.8023 – 0.2912
= 0.5111
Application of Standard Normal
Distribution
Q3: In an industrial process, the diameter of a
ball bearing is an important measurement.
The buyer sets specifications for diameter to
be .3.0  0.01 cm The implication is that no part
falling outside these specifications will be
accepted. It is known that in the process the
diameter of a ball bearing has a normal
distribution with mean = 3.0 and standard
deviation = 0.005. On average, how many
manufactured ball bearing will be scrapped?
Application of Standard Normal
Distribution
Solution:
2.99  3.0
z1 
 2.0
0.005
3.01  3.0
z2 
 2.0
0.005
Application of Standard Normal
Distribution
• P (2.99 < X < 3.01) = P (-2.0 < Z < +2)
• P (-2.0<Z)= 0.0228
• P (Z < +2) = 1 - P (+2<Z) = 1- 0.9772 =
0.0228
• Then the average that is 2  0.0228  0.0456
So that 4.56% are scrapped
Application of Standard Normal
Distribution
Q4: Gauges are used to reject all
components for which a certain
dimension
is
not
within
the
specification. It is known that this
measurement is normally distribution
with mean 1.50 and standard deviation
0.2. Determine the value d such that the
specifications cover 95% of the
measurements.
Application of Standard Normal
Distribution
The left area = 100 – 95
= 5% for both sides
0.05
In each side 
 0.025
2
Application of Standard Normal
Distribution
• The area is equal to 1 – 0.025 = 0.975
• From table this area equal to z1 or z2 =
1.96
(1.50  d )  1.5
0.2
1.96  0.2  1.5  1.5  0.392  d
1.96 
Or
(1.50  d )  1.5
1.96 
0.2
1.96  0.2  1.5  1.5  0.392   d
Application of Standard Normal
Distribution
Q5: A certain machine makes electrical
resisters having mean resistance of 40
ohms and standard deviation of 2.0
ohms. Assuming that the resistance
follows a normal distribution and can
be measured to any degree of accuracy,
what percentage of resistors will have
resistance exceed 43 ohms?
Application of Standard Normal
Distribution
Solution:
34  40
z
 1.5
2
P (X > 43) = P(Z > 1.5) = 1- P(Z < 1.5)
=1 – 0.9332=0.0668
Application of Standard Normal
Distribution
• Solution:
• Then the percentage of resistors will
have resistance exceed 43 ohms =
6.68%
Application of Standard Normal
Distribution
Q6: The average grade of an exam is 74%
and the standard deviation is 7.0. If 12%
of the class is given A’s and the grades
are
curved
to
follow
normal
distribution, what is the lowest possible
A and the highest possible B?
Application of Standard Normal
Distribution
Solution:
The area left 12% is 88% = 0.88
From table we found the z value is 1.18
Application of Standard Normal
Distribution
• Then
z 
x

x  74
1.18 
7
 x  82.26
Lowest A is 82 and highest B is 83