WS3_Worksheet_Solutions(1)
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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
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CHEE2010 Workshop 3 2024: Z, α, percentiles and probabilities for normal distributions
the resistance of any given resistor will exceed 10.05 ohms? [
this could also be written: what proportion of resistors do you expect to find with R>10.05 ohms?]
Z
X
=
10.05
=
10.05
−
10.02
0.19
=
0.158
P(Z≤0.158) = 0.563
Therefore P(Z>0.158) = 1- P(Z≤0.158) = 1-0.563 = 0.437 = 43.7%
c.
Assuming that the mean of the resistance cannot be changed due to process constraints, what change in the standard deviation of the population would ensure that 99.9 % of the resistors satisfy R≤ 10.05 ohms?
α = proportion of data above the percentile = 1- proportion of data below the percentile
α = 1-0.999 =0.001
Z
X
= Z
α=0.001
= 3.09 (by Excel norm.inv(0.001,0,1)) or, using the Z table: P(X > x) = P(X <= -x)
P
(
Z≤Z
X
)
= 0.001, therefore P((Z > -
Z
X
¿
= 0.001, P
(
Z≤
−
3.09
)
= 0.001, therefore P((Z > +
3.09
¿
= 0.001
Note: Z table may give you 3.08 or 3.09. This is fine.
3.09
=
10.05
−
10.02
α
α
=
0.03
3.09
=
0.09
ohms
3.
A manufacturing plant knows that it typically produces steel pipes with a mean diameter (µ) of 123 mm and a standard deviation (σ) of 5 mm. If you were to take a sample of 100 pipes, what is the expected mean with associated 95% confidence interval for that sample?
Z-crit: +/-1.96
Standard error = s
µ
=
σ
√
n
=
5
mm
√
100
=
0.5
mm
E
95 = s
µ
∗
Z
crit
= 0.5 mm * 1.96 = 0.98 mm = 1.0 mm
Expected Mean = 123 +/- 1 mm for the sample (note cannot report to more than whole numbers based on the initial precision of measurement)
CHEE2010 Workshop 3 2024: Z, α, percentiles and probabilities for normal distributions
4.
Twenty paper samples are taken from two different manufacturing sites (with both sets of samples being taken at the same time). The first set of twenty samples (location 1) returns an observed mean density of 1.032 g/mL with an observed standard deviation of 0.015 g/mL, while the second set of twenty samples (location 2) returns an observed mean density of 1.018 g/mL with an observed standard deviation of 0.012 g/mL. Determine the value and uncertainty (95% confidence interval) of the paper density from
each location. Based on location and spread of the confidence intervals, would you conclude that there are differences in the two means? Degree of freedom =n-1 =20-1 = 19
We have been given that we have a 95% confidence interval, which means that α =0.05, in other words we assume that 5% of the data is not within the range of the confidence interval. Because this is a two-tailed question, you have 2.5% on the right hand side that is not within the range of the confidence interval and 2.5% on the left hand side that is not within the range of the confidence interval. Therefore the upper value of t
crit is set so that 97.5% of the data is below this value. Therefore we want to find t
0.975,df
t
crit
: in Excel t.inv(0.975,df) = t.inv(0.975,19). Check the t table
t
crit
= t
0.975,19
= 2.093
Standard error = s
X
=
s
√
n
For sample 1, s
X
=
0.015
√
20
=
0.00335
For sample 2, s
X
=
0.012
√
20
=
0.00268
E
95
= Margin of error = t
crit
* s
X
E
95,1
= 2.093*0.00335 = 0.007. Mean = 1.032 ± 0.007 g/mL
E
95 2 = 2.093*0.00268 = 0.006 Mean = 1.018 ± 0.006 g/mL
CIs do not overlap – there is a difference in means