Probability and Data Analysis

This assignment delves into the concepts of probability and data analysis. It presents four quizzes with skewed data distributions (leftward and rightward) and explores the average performance across each quiz. It then focuses on a scenario involving the probability of an alternator failure during a flight, requiring calculations for both failures and non-failures. Lastly, it examines a dataset of steps recorded over time, calculating a 95% confidence interval and determining the sample size needed to achieve a specific margin of error. The assignment concludes with a line chart interpretation.

sta101 STATISTICS FOR BUSINESS
Assignment
Student id
[Pick the date]

Question 1
S.
No.
Question Data Type Measuremen
t level
Reasoning
1. Gender Categorical Nominal The reason is that gender i.e.
being male and female would be
measured on nominal scale.
2. Approximate
undergraduate college
GPA
Continuous Ratio Undergraduate college GPA i.e.
1 to 4 is measured on ration
scale and is of continuous data
type.
3. Number of hours per
week
Continuous Ratio Number of hours per week
would be a numeric value and
hence, solved through
mathematical operation.
Therefore, the data type is
continuous and has measured at
ratio measurement scale.
4. Ideal number of children
for a married couple
Discrete Ratio Ideal number of children would
be integer value and hence,
solved through mathematical
operation. Therefore, it has
measured at ratio measurement
scale.
5. Describe parents Categorical Ordinal Order for one month would also
dominate up through order 5 for
the father. Therefore, , it has
measured at ordinal
measurement scale.
1

Question 2
Relevant data is represents below:
(a) The mean, median and mode of the above data is computed through excel functions and
is shown below:
Function used:
Mean = AVERAGE (), Median = MEDIAN (), Mode = MODE ()
(b) The measures of central tendency (mean, median and mode) are not equal and therefore,
it would be fair to conclude that the measures of central tendency do not agree.
(c) The standard deviation has been calculated through excel function and is shown below:
2

Function used:
STDEV ()
(d) The sorting and standardized data is represented below:
zi= ( xi−x
s )
x=724.67, s=114.28, xi is given set of data
3

(e) Unusual values fall when the values are out of x ± 2 s range and the outliers would present
when the values are out of x ± 3 s range .
For Outliers
x ± 3 s=724.67 ± ( 3∗114.3 )
x +3 s=724.67+ ( 3∗114.3 )=381.8
4

x−3 s=724.67− ( 3∗114.3 ) =1067.6
It can be seen that all the provided values are fall between this range and therefore, the data does
not has any outliers.
For unusual observation
x ± 2 s=724.67 ± ( 2∗114.3 )
x +2 s=724.67+ ( 2∗114.3 )=496.1
x−2 s=724.67− ( 2∗114.3 )=953.3
It can be seen that from the provided values only observation number 1030 does not fall between
ranges and therefore, the data has unusual observation.
(f) It can be concluded based on the “three sigma empirical rule” that the provided data are
derived from normal population.
The main reason behind this understanding is that approximately 99.7% of provided data lie
within in the domain of three standard deviation of mean.
Three standard deviation of mean would be determined as shown below:
x ± 3 s=724.67 ± ( 3∗114.3 )
x +3 s=724.67+ ( 3∗114.3 )=381.8
x−3 s=724.67− ( 3∗114.3 ) =1067.6
Range[381.8 1067.6]
5

Question 3
(a) Mean, median and mode calculation of quiz
For quiz 1
Quiz 1 :60 , 60 , 60 ,60 , 71 ,73 , 74 , 75 , 88 , 99.
Total count=10
Mean=((60+60+ 60+60+71+73+74+ 75+ 88+99)/10)=720/10
Mean=72
Median=((71+73))/2=72
Mode=6 0
For quiz 2
Quiz 2 :65 , 65 , 65 ,65 , 70 , 74 , 79 ,79 , 79 ,79
Total count=10
Mean=¿(65+65+65+65+70+74+79+79+79+79)/10=720/10=72
Median=¿(70+74)/2¿ 72
Mode=79∧65
For quiz 3
Quiz 3: 66, 67, 70, 71, 72, 72, 74, 74, 95, 99
Total count=10
Mean=¿(66+67+70+71+72+72+74+74+95+99)/10=760/10=76
Median=(72+72)/2=72
Mode=7 4∧72
For quiz 4
Quiz 4: 10, 49, 70, 80, 85, 88, 90, 93, 97, 98
Total count = 10
6

Mean =(10+49+70+80+85+88+90+93+97+98)/10=760/10=76
Median=(85+88)/2=86.50
Mode= No mode as there is no repetitive frequency for any data point.
(b) The measures of central tendency (mean, median and mode) are not equal and therefore, it
would be fair to conclude that the measures of central tendency do not agree.
(c) The mean, median for the quizzes has been determined. Further, there is no mode for quiz 4.
(d) The data of quiz 1 represents leftward skew.
The data of quiz 2 represents symmetric.
The data of quiz 3 represents rightward skew.
The data of quiz 4 represents leftward skew.
(e) The average performance for the quiz 1 is 72.
The average performance for the quiz 2 is 72.
The average performance for the quiz 3 is 76.
The average performance for the quiz 4 is 76.
Question 4
Computation of probabilities
Probability that the alternator would fail on a 1hr flight = 0.2
7

(a) The requisite probability that both would fail
P ( bothwill fail ) =0.02∗0.02=0.000 4
(b) The requisite probability that neither would fail
P ( neither will fail ) =¿
(c) The requisite probability that one or other would fail
P ( one∨other would fail )=1−P ( neither will fail )=1−0.9604=0.0396
Question 5
(a) The value of 95% confidence interval is computed below:
The value of mean = 3,279
The value of standard deviation = 97.05
The number of observation (sample size) = 18
Degree of freedom = 18 – 1 = 17
In the accordance of central limit theorem, it can be seen that sample size is lower than 30 and
therefore, z value would not be used for the computation of confidence interval. Hence, t value
would be taken into consideration.
t value for17 degree of freedom and for 95% confidence interval comes out to be 2.1098.
95 % confidence interval=Mean ±¿
 Lower limit of 95% confidence interval
8

¿ Mean−¿
= 3279−2.1098 ( 97.05
√18 )
¿ 3230.4 5
 Upper limit of 95% confidence interval
¿ Mean+¿
= 3279+2.1098 ( 97.05
√18 )
¿ 3326.9 8
Therefore, 95% confidence interval is 3230.45 3326.98.
(b) The objective is to determine the number of observation or sample size in regards to obtain
error of ± 20 steps with the 95% confidence interval.
Margin of error = t value * (standard deviation) / sqrt (sample size)
Margin of error = 20
Standard deviation = 97.05
Sample size =?
Margin of error = t value * (standard deviation) / sqrt (sample size)
20=2.1098∗( 97.05 )
√¿ ¿
√ ¿
Sample size = 104.81 or 105
9

Hence, the requisite sample size would be 105.
(c) The line chart of the data is represented below:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
2,900
3,000
3,100
3,200
3,300
3,400
3,500
Line Chart
Time (Days)
Distance recorded (miles)
“From the above shown line chart, it would be fair to conclude that distance recorded i.e. number
of steps would fall with respect to the time.”
10

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