WS3_Worksheet_Solutions(1)
.docx
keyboard_arrow_up
School
The University of Queensland *
*We aren’t endorsed by this school
Course
2010
Subject
English
Date
Apr 3, 2024
Type
docx
Pages
5
Uploaded by GrandLightning11612 on coursehero.com
CHEE2010 Workshop 3 2024: Z, α, percentiles and probabilities for normal distributions
1.
Making sense of Z, α, percentiles and probabilities for normal distributions
Any data X which is normally distributed with mean µ and standard deviation σ can be converted to the standardized normal distribution
as follows: Z=(X- µ
x
)
/ σ
x
a.
Calculate the mean of Z: μ
Z
=
X
−
μ
X
σ
=
X
σ
−
μ
X
σ
≈
μ
X
σ
−
μ
X
σ
=
0
b.
For the standardize normal distribution, 34 % of the data lies within one standard deviation below the mean, and 34 % of the data lies within one standard deviation above the mean: draw this on a simple probability density figure, where the y axis is probability, and x axis is value of Z
CHEE2010 Workshop 3 2024: Z, α, percentiles and probabilities for normal distributions
c.
Complete the following tables for each of the percentiles defined below. There is enough data in the table to complete all the missing information. To calculate other z
α
values, using command in Excel =norm.inv(Proportion of data which falls below this percentile,0,1)
Table 1: Percentiles, z scores and α values for a standardized normal distribution Percentile
25
50
75
84
90
95
97.5
99
99.5
99.9
Proportion of data which falls below this percentile
0.25
0.5
0.75
0.84
0.9
0.95
0.975
0.99
0.995
0.999
z
α
=number of standard deviations above the mean corresponding to the percentile
-0.6745
0
0.6745
1
1.28
1.645
1.96
2.33
2.58
3.08
P(Z> z
α
)= α
0.75
0.5
0.25
0.16
0.1
0.05
0.025
0.01
0.005
0.001
P(|Z |< | z
α
|), i.e. The probability that Z lies within distance of z
α
from
the mean
0.5
0
0.5
0.68
0.8
0.9
0.95
0.98
0.99
0.998
For the last row, this is equivalent to saying P(|Z |< | z
α
|) =1-2α
CHEE2010 Workshop 3 2024: Z, α, percentiles and probabilities for normal distributions
2.
Determining information from normally distributed populations The resistance R of a batch
of resistors is normally distributed. Testing has shown that 10 % of the resistors have R>10.256 ohms, and 5 % have R<9.710 ohms. a.
What is the mean and standard deviation of resistance R for the population of resistors?
Assume resistance is normally distributed
Remember Z
=
X
−
μ
X
σ
Percent resistors with R≤10.256 ohms =100-10 =90% Percent resistors with R>10.256 ohms = 10% α = 0.1. P(Z ≤ Z
X=10.256
) = 0.90 Z
X=10.256
= Z
α=0.1
= 1.286
Z
X=10.256
=
Z
0.10
=+
1.286
=
10.256
−
μ
X
σ
10.256
−
1.286
σ
=
μ
X
Percent resistors with R≤9.710 ohms = 5% Percent resistors with R>9.710 ohms = 95% α = 0.95. P(Z ≤ Z
X
) = 0.05
Z
0.95
: remember Z
0.95 is the value of Z (the number of standard deviations above or below the mean) at which 95% of the data lies above this value for a standard normal distribution. Also since the normal distribution is symmetric, then Z
α = - Z
1-α
So Z
X=9.710
= Z
α=0.95 = - Z
α=0.05
= -1.645
Z
X
=
9.710
=
Z
α
=
0.95
=−
1.645
=
9.710
−
μ
X
σ
μ
X
=
1.645
σ
+
9.71
Substituting in from above
10.256
−
1.286
σ
=
1.645
σ
+
9.71
0.546 = 2.931
σ
σ
=
0.546
2.931
=
0.186
ohms
………..using correct sig figs = 0.19 ohms
Then substitute back in to one of the Z calculations
Z
X
=
9.710
=
Z
α
=
0.95
=−
1.645
=
9.71
−
μ
X
0.186
(
−
1.645
∗
0.186
)
−
9.71
=−
μ
X
μ
X
=
10.016
ohms
Using correct sig figs based on σ: μ
X
=
¿
10.02 ohms
b.
The resistors are to be used in an application where the resistance is consistently under 10.05 ohms. Under the current circumstances, what is the probability that
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help