Statistics Assignment: Probability Analysis and Data Interpretation

This statistics assignment delves into various statistical concepts and techniques. It begins with an introduction to probability and random variable analysis, including examples related to car WOF certifications and rolling dice. The assignment then explores probability distributions, specifically binomial distributions, and analyzes the probability of sample variables. It covers scenarios such as analyzing the probability of people participating in a stop smoking program and calculating probabilities related to car speeds. Further, the assignment includes calculations of Z-scores and analyzes probabilities using the standard normal distribution. Finally, the assignment addresses hypothesis testing, comparing the mean value of life shocks and stress on families and includes a detailed analysis of the data with statistical measures like mean, standard deviation, and confidence levels, providing a comprehensive overview of statistical analysis.

STATISTICS

TABLE OF CONTENTS
INTRODUCTION...........................................................................................................................1
QUESTION 1...................................................................................................................................1
(a). Random variable interest analysis on randomly selected cars with Warrant of Fitness
(WOF).....................................................................................................................................1
(b). Identification of specific data collection..........................................................................1
(c). Determining the probability of having double 6 with rolling 2 dice...............................2
QUESTION 2...................................................................................................................................2
(a) Analysing the probability distribution of X......................................................................3
(b) Analysing the probability of sample variables.................................................................3
1).............................................................................................................................................3
2).............................................................................................................................................3
3).............................................................................................................................................4
4).............................................................................................................................................4
(c) Analysing the probability of the people which goes under the program of stop smoking4
1).............................................................................................................................................4
2). Success rate of this new programme is higher than 50%..................................................5
3). Justification about actual success rate higher than 65%...................................................5
QUESTION 3...................................................................................................................................5
A. Cars expected to be travel slower than 60 km/h................................................................5
B. Cars expected to have speed exceeding the speed limit....................................................5
C. Car's speed faster by 5%....................................................................................................6
D. Reduction in the car speed by 5%.....................................................................................6
E. Measuring standardise value..............................................................................................6
QUESTION 4...................................................................................................................................7
4.A..........................................................................................................................................7
4.B..........................................................................................................................................7
4.C..........................................................................................................................................7
4.D..........................................................................................................................................8
4.E...........................................................................................................................................8

QUESTION 5...................................................................................................................................8
5(a)..........................................................................................................................................8
5(b).........................................................................................................................................9
QUESTION 6.................................................................................................................................10
(a)..........................................................................................................................................10
1). 90% of confidence interval.............................................................................................10
2). 95% of confidence level..................................................................................................11
(b).........................................................................................................................................11
(c)..........................................................................................................................................12
(d).........................................................................................................................................12
QUESTION 7.................................................................................................................................12
(a) Analysing the increment in the height of students..........................................................12
(b) Implication of test...........................................................................................................12
CONCLUSION..............................................................................................................................13
REFERENCES..............................................................................................................................14

INTRODUCTION
To analyse a financial statement, it is necessary to consider all techniques performed by
influencing statistical measurement. Implicating various tools and techniques which will be
helpful in determining probability of variables which will be adequate in determining concrete
solutions. In the present report there will be use of various methods such as random variable
analysis, SPSS tools, Z score techniques etc. in analysing data base. These outcomes will be
helpful in making adequate operational control and determination of all variables in a right
manner.
QUESTION 1
(a). Random variable interest analysis on randomly selected cars with Warrant of Fitness (WOF)
No. of cars 30
probability of car have WOF 1/6
Random variable P(X=0) = 1/6
X 30
P(30=0) = 1/6
P 0.2
0≤ P(X)≤ 1
0≤ 0.2≤ 1
Interpretation: By considering operational variations in the outcomes which identifies
that there will be determination of adequate functions. There has been selection of 30 random
cars with the motive of examining WOF certification obtained by them in practice. Along with
this, as per considering the probability obtaining certificates by these cars which is 1/6 (Pope and
Stanistreet, 2017). Thus, as per analysing P(X) value there has been determination of all relevant
outcomes such as 0.2 on the basis of formula P (30=0) = 1/6.
(b). Identification of specific data collection
No. of responses 100
1

Probability of selected adults in New Zealand 0.40
0≤ 0.4≤ 1
P 0.4
Random variable P(X=0) = 0.4
P (100=0) = 2/5
Probability of selected adults in New Zealand 2/5
Interpretation: As per analysing above listed outcomes which will be adequate and
helpful as per making quantitative determination of all operations in right manner. The number
of responses which have been studied and analysed here are 100 respondents. The P(X) is of 0.40
(Statistical data analysis, 2018). However, this is a reverse approach which in turn will be used
for analysing the fractional value of probability. Thus, with influences of formula such as P
(100= 0) = 0.4, which defines probability of 2/5.
(c). Determining the probability of having double 6 with rolling 2 dice
Two Dice 6+6
Chance of having Double Six 6/36
Probability of double 6 1/6
Interpretation: In relation with identifying the probability of having double 6 with
rolling 2 dice there will be probability of this combination is 6/36. Moreover, as per resolving
this fraction which indicates outcomes as 1/6. Thus, it can be said that there will be chances of
having double 6 from 2 dice are 1/6.
QUESTION 2
success rate 0.55
Failure rate 0.45
n 100
2

(a) Analysing the probability distribution of X
Binomial distribution
Justification There is set of n trials are conducted
There is success or failure
P(X=0.55)
Interpretation: As per analysing the outcomes which determines that there will be
identification of probability distribution of X with implicating the 'X~' Notation. However, in
relation with obtaining the results with the influences of set of N trials which have been
conducted and that presents P (X= 0.55) (Cho and et. al., 2015).
(b) Analysing the probability of sample variables
b.
Let X be the number of
people who manage to
stop smoking
1) X 50
n 100
0.0481519715
Interpretation: By considering, above listed measurements and determining the
probability of 50 people who have managed to quit smoking. Therefore, the X value of data set
can be determined as 50 who are quitting smoking out of N number of population which is 100.
However, as per analysing probability of people who wanted to quit smoking such as 0.048151.
2).
X Less than 50
n 100
P(X<50) 0.1827281847
3

Interpretation: In relation with analysing probability of people fewer than 50 individuals
who are willing to stop smoking. Thus, it identifies the X value of less than 50, N as 100 which
had been put into formula of P(X<50) presents answer as 0.182.
3)
X more than 50
n 100
1-P(X<50) 0.8172718153
P(X>50) 0.8172718153
Interpretation: By analysing the probability of people who are more than 50 individuals
as per stop smoking has been identified. Thus, as per analysing the outcomes there has been use
of various techniques which in turn will be useful for making better operational gains. X variable
denotes the people who are more than 50, N as 100, 1-P(X<50) as 0.8173 and P(X<50) as
0.81727.
4).
X 90-100
n 100
Bino midst range 0
Interpretation: On the basis of above listed analysis it can be said that there have been
90-100 people which engaged in program of stop smoking. Therefore, the Bino midst range of
the outcomes has been gathered which is 0.00000025 approximately 0.
(c) Analysing the probability of the people which goes under the program of stop smoking
1).
X= or more than 68 68
n 100
4

p 0.5
P(X<68) 0.99991
P(X>68) 1 – P(X<68)
P(X>68) 0.01
Interpretation: In context with the above listed analysis where 68 people has managed
to stop smoking out of 100. Thus, X will be 68, N as 100 and P as 0.5. Thus, as per analysing the
outcomes there will be implications of formula P(X<68) = 0.99, P(X>68) = 1-P(X<68) and
P(X>68) = 0.01.
2). Success rate of this new programme is higher than 50%
The outcome which is giving success rate as 50% with higher probability as 0.01 but on
its contrary, its probability of less than 68 is higher than compared to other. Its success rate is
less than 68 due to high probability.
3). Justification about actual success rate higher than 65%
Yes, it could be justified by observing probability of both more and less than 68. There
are more chances (0.99) for getting success rate as less than 50%.
QUESTION 3
A. Cars expected to be travel slower than 60 km/h.
proportion of cars will be slower than 60 km/h
mean f/Ƹf
69 60/Ƹf
Ƹf 69*60
Ƹf 4140
Interpretation: In order to consider the cars which will travel slower than 60 km/h can
be determined as per analysing it with mean speed of cars such as 69 km/h. However, as per
analysing the submission value of outcomes which is 4140.
B. Cars expected to have speed exceeding the speed limit
Proportion of cars breaking speed limit 75 km/h
mean f/Ƹf
5

69 75/Ƹf
Ƹf 69*75
Ƹf 5175
Interpretation: As per considering the outcomes which identifies cars which have
crossed speed limit of 70 km/h. Thus, it has been estimated as 75 km/h and the submission value
will be 5175.
C. Car's speed faster by 5%
Speed of the fastest car will be exceeding 5%
With average speed limit 69 km/h
exceeding speed will be 72.5
Interpretation: In relation with analysing the speed of cars which will rise by 5%. Then
there will be rise in the average speed limit of 69 km/h. Thus, the average will be rise to 72.5
km/h.
D. Reduction in the car speed by 5%
Speed of car will be reduced 5%
With speed limit 69 km/h
Reducing speed will be 65.5
Interpretation: By considering the reduction in car speed by 5% which will be analysed
on determining average speed of 69 km/h. Thus, there has been reduction in average speed limit
of around 65.5 km/h.
E. Measuring standardise value
x 70 km/h 79.5
Z score value
Z (X- μ)/ 𝞼 (X- μ)/ 𝞼
Mean μ 69km/h 69km/h
6

Standard deviation 𝞼 3.5 km/h 3.5 km/h
(70-69)/3.5 (79.5-69)/3.5
Z score 0.285714 3
Interpretation: In relation with analysing Z score value of outcomes on the basis of
speed limit 70 km/h and 79.5 km /h. However, as per implicating value under formula which
identifies the outcomes as (X-μ) / 𝞼. Thus, in 70 km/ h there will be Z score of 0.285 and in 79.5
km/h it is 3 (SPSS, M.O.D., 2015).
QUESTION 4
4.A
P(Z<0) = 0.5
4.B
P(Z>2) 1- P(Z<2)
P(Z<2) 0.98
P(Z>2) 0.23
Interpretation: The above tables are representing probabilities by considering properties
of standard normal random variable Z. The probability of Z which is less than zero is stated as
0.5. Further it had reflected probability of getting Z as more than 2 which is 0.23.
4.C
P(IZI < 2) P(Z > -2) + P(Z < 2) 1.44333987
P(Z > -2) 0.46609 1 – P(Z < -2)
P(Z < 2) 0.97724987
7

4.D
P(IZI > 2) P(Z < -2) + P(Z > 2) 0.04550026
P(Z < -2) 0.02275013
P(Z > 2) 0.02275013
Interpretation: The absolute value of z as less than 2 but it is not negative which gives
probability of 1.4433. In the same series, it is giving absolute value of more than 2 as 0.045
(Park, Yu and Jo, 2016).
4.E
P(-3 < z < 3) P(Z < +3) – P(Z < -3)
P(Z < +3) 0.9986501
P(Z < -3) 0.0013499
P(-3 < z < 3) 1
QUESTION 5
Hypothesis 0: There is no Significant differences between the mean value of Life shocks and
Stress on families.
Hypothesis 1: There is a significant difference between the mean value of Life shocks and stress
on families.
5(a)
Random selected families
Mean 3.166666667
Standard Error 1.085952545
Median 1.5
Mode 1
Standard Deviation 3.761849964
Sample Variance 14.15151515
Kurtosis 1.528517257
Skewness 1.430568639
Range 12
8

Minimum 0
Maximum 12
Sum 38
Count 12
Confidence Level(95.0%) 2.390165435
5(b)
Random selected families
0
0
0
1
1
1
2
3
4
6
8
12
Calories
Q1 0.75
Median - Q1 0.75
Q3-median 3
mean 3.17
9