Statistical Analysis and Probability Assignment 1 Solution

This document presents a comprehensive solution for a statistics assignment, addressing two main tasks. Task 1 involves calculating the mean and standard deviation from a set of calibration data, analyzing grouped data, determining the probability of a disk diameter falling within a specified tolerance, and calculating the product moment correlation coefficient and regression line. Task 2 focuses on the application of binomial distribution to determine the probability of disks being out of tolerance in a pack, as well as calculating the probability of a disk's diameter exceeding a certain value and falling within a range. The solution provides step-by-step calculations, tables, figures, and relevant formulas, demonstrating a thorough understanding of statistical concepts and their practical application. References are also included.

Running head: SOLUTION FOR ASSIGNMENT 1
Solutions for Assignment
Name
Institution

SOLUTION FOR ASSIGNMENT 2
Solution for Assignment
Task 1
a) Let the calibration be represented by x. The mean ( X ) and standard deviation ( s ), by
definition, are given using the formula:
X =
∑
i=1
n
X
n
and s= √ ∑
i=1
n
( X −X ) 2
n−1
Where:
n = sample size (30).
∑
i=1
n
X = sum of all the calibrations.
∑
i=1
n
( X− X ) 2 = sum of the square difference between calibration and mean.
Using excel sum function ∑
i=1
30
X=7201.05. Thus,
X =7201.05
30 =240.035mm.
In order to get the results for s, the calculations are shown in table below.
X X X
239.73 0.09 240.14 0.01 239.91 0.02
239.81 0.05 239.99 0.00 240.23 0.04
240.58 0.30 240.80 0.59 240.06 0.00
240.10 0.00 240.23 0.04 239.74 0.09
240.87 0.70 240.11 0.01 240.01 0.00
240.24 0.04 240.08 0.00 239.95 0.01
239.97 0.00 239.78 0.07 240.23 0.04
240.16 0.02 240.43 0.16 239.60 0.19
239.28 0.57 239.01 1.05 239.94 0.01
( X −X ) 2
( X −X ) 2
( X −X ) 2

SOLUTION FOR ASSIGNMENT 3
240.07 0.00 239.69 0.12 240.31 0.08
Sum 1.78 2.03 0.46
∑
i=1
n
( X− X ) 2
4.27
The standard deviation is s= √ 4.27
30−1 =0.38 mm
b) The table below shows the grouping of the data
Diameter
(mm)
239.0 ≤ d < 239.4 239.4 ≤ d < 239.8 239.8 ≤ d < 240.2 240.2 ≤ d < 240.6 240.6 ≤ d < 241.0
Tally / //// / //// //// //// //// // //
Frequency 1 6 14 7 2
c) The figure 1 and 2 shows barg graph and ferquency polygon for the data.
239.0 - 239.4 239.4 -239.8 239.8 -240.2 240.2 - 240.6 240.6 - 241.0
0
2
4
6
8
10
12
14
16
Figure 1: Bar Graph of Diameter (mm)
Diameter (mm) class
Frequency

SOLUTION FOR ASSIGNMENT 4
239.0 - 239.4 239.4 -239.8 239.8 -240.2 240.2 - 240.6 240.6 - 241.0
0
2
4
6
8
10
12
14
16
Figure 2: Frequency Polygon for Diamter (mm)
Diameter (mm) Class
Frequency
d) For grouped data, mean ( X ) and standard deviation ( s ), by definition, are given using the
formula (Wilson, 2016):
X =∑ fX
∑ f and s= √ ∑ f ( X−X )2
∑ f −1
Where:
X = is the mid-point of the class
f = class frequency
The table below shows the preliminary calculations
x f fx ( X −X ) 2 f ( X−X ) 2
239.2 1 239.2 0.71 0.71
239.6 6 1437.6 0.19 1.16
240.0 14 3360 0.00 0.02
240.4 7 1682.8 0.13 0.91
240.8 2 481.6 0.58 1.16
∑ 30 7201.2 3.952
From table above, ∑ fX =7201.2 ,∑ f =30 ,∧∑ f ( X−X )2=3.952.
Then, X =7201.2
30 =240.04 mm and s= √ 3.952
30−1 =0.34mm.

SOLUTION FOR ASSIGNMENT 5
e) The disk is within tolerance if the diameter is 239.40 ≤ d ≤ 240.60.
The, the probability that the diameter is within tolerance is given by:
Prob ( Tolerance ) =P ( 239.40 ≤ d ≤ 240.60 )
The, probability that the disk is out of tolerance is:
Prob ( Out of Tolerance )=1−P ( 239.40 ≤ d ≤ 240.60 )
Since, sample size is large (30) we assume normality and use normal standardized tables
to obtain the probability as follows:
√ n ( X−X )
s =Z N (0 , 1)
Now,
P ( 239.40≤ X ≤ 240.60 )=P ( √ 30 ( 239.40−240.04 )
0.34 ≤ √30 ( X−240.04 )
0.34 ≤ √30 ( 240.60−240.04 )
0.34 )
P ( 239.40≤ X ≤ 240.60 )=P (−10.31≤ Z ≤ 9.02 )
P ( 239.40≤ X ≤ 240.60 ) =Φ ( 9.02 ) −Φ ( −10.31 )
1.2
1.0
0.8
0.6
0.4
0.2
0.0
X
Density
239.4
0.9203
240.6240.0
Distribution Plot
Normal, Mean=240.04, StDev=0.34
P ( 239.40≤ X ≤ 240.60 )=0.9203
Then,
Prob ( Out of Tolerance )=1−0.9203=0.0797

SOLUTION FOR ASSIGNMENT 6
f) The mean and standard deviation is given as follows:
X =∑ X
n and s= √ ∑ ( X −X ) 2
n−1
Lathe No.
No. Produced
(x)
No. out of
Tolerance (y)
1 102 4 31.36
2 98 3 2.56
3 88 1 70.56
4 104 6 57.76
5 91 3 29.16
6 97 5 0.36
7 101 5 21.16
8 93 4 11.56
9 94 2 5.76
10 96 3 0.16
∑ 964 230.4
From table above, ∑ X=964 and ∑ ( X− X )2=230.4 , n=10
X = 964
10 =96.4 disks and s= √ 230.4
10−1 =4.8 disks.
g) The product moment correlation coefficient and regression line are calculated in excel as
follows (Rodgers, 2016):
product moment correlation coefficient = 0.7964.
Regression line:
y=−19.245+ 0.237 x
Where:
y=Number of disks out of tolerance
x=number of disks produced
h) The scatter plot with regression line and equation produced in excel is presented below:
( X −X ) 2

SOLUTION FOR ASSIGNMENT 7
86 88 90 92 94 96 98 100 102 104 106
0
1
2
3
4
5
6
7
f(x) = 0.236979166666667 x − 19.2447916666667
Scatter Plot of Y agains X
Number of disks produced (x)
Number of disks out of tolerance (y)
Task 2
(a) Given P0 oT =0.022, packed of eight, and assumed binomial distribution. Let x represent
the number of disks out of tolerance. Then x is assumed to binomially distributed:
x bin(n , p) where: n = 8 and p = P0 oT =0.022.
Pr ( X=x )= (8
x )0.022x ( 1−0.022 )8− x
Pr ( X=x ) = ( 8
x ) 0.022x ( 0.9788− x )
(i) The probability that in a pack of 8 there is no disks out of tolerance is given as
Pr ( X=0 )=(8
0 )0.0220 ( 0.9788 −0 )
Pr ( X=0 ) =0.9788
Pr ( X=0 ) =0.8370
(ii) The probability that at most one disk is out of tolerance
Pr ( At most one out of tolerance )=Pr ( X ≤ 1 )

SOLUTION FOR ASSIGNMENT 8
Pr ( X ≤ 1 ) =Pr ( X=0 ) + Pr ( X =1 )
But, from (i) Pr ( X=0 ) =0.8370
Pr ( X=1 ) = ( 8
1 ) 0.0221 ( 0.9788−1 )
Pr ( X=1 )=8(0.022)(0.9787 )
Pr ( X=1 )=0.1506
Then,
Pr ( X ≤ 1 )=0.8370+0.1506
Pr ( X ≤ 1 ) =0.9876 .
(b) Given X =240.15 and s=0.4
(i) The probability that the disk has diameter over 240.7mm is given as
follows:
Pr ( Diameter
240.7 mm )=Pr ( X >240.7 )
Since we have assumed normality we then standardize and use standard
normal tables as follows:
Pr ( X >240.7 )=1−Pr ( X ≤ 240.7 )
Pr ( X >240.7 )=1−Pr ( ( X−240.15 )
0.4 ≤ ( 240.7−240.15 )
0.4 )
Pr ( X >240.7 ) =1−Pr ( Z ≤ 1.375 )
Pr ( X >240.7 )=1−Φ ( 1.375 )
Pr ( X >240.7 )=1−0.9154
Pr ( X >240.7 )=0.0846

SOLUTION FOR ASSIGNMENT 9
1.0
0.8
0.6
0.4
0.2
0.0
X
Density
240.7
0.08457
240.2
Distribution Plot
Normal, Mean=240.15, StDev=0.4
(ii) The probability that a disk has a diameter of between 239.6 and 240.6
Pr ( diameter between 239.6∧240.6 )=Pr ( 239.6 ≤ X ≤ 240.6 )
Pr ( 239.6 ≤ X ≤240.6 ) =Pr ( ( 239.6−240.15 )
0.4 ≤ Z ≤ ( 240.6−240.15 )
0.4 )
Pr ( 239.6 ≤ X ≤240.6 ) =Pr (−1.375≤ Z ≤1.125 )
Pr ( 239.6 ≤ X ≤240.6 )=Φ ( 1.125 ) −Φ (−1.375 )
Pr ( 239.6 ≤ X ≤240.6 )=0.8697−0.0846
Pr ( 239.6 ≤ X ≤240.6 ) =0.7851
1.0
0.8
0.6
0.4
0.2
0.0
X
Density
239.6
0.7851
240.6240.2
Distribution Plot
Normal, Mean=240.15, StDev=0.4

SOLUTION FOR ASSIGNMENT 10
References
Rodgers, K. A. (2016). Correlation Analysis with Excel Handout.
Wilson, K. (2016). Essential Excel 2016. Elluminet Press.