Rahul_Kaushal_Case_Study
xlsx
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School
Centennial College *
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Course
4001
Subject
Medicine
Date
Apr 3, 2024
Type
xlsx
Pages
28
Uploaded by GeneralSteelPony25
rking the Case Stud
Case Study Tab Name
Marks
Question Number
Breakdown of Marks
Housing Price
5
a
1.5
b
1.5
c
0.5
d
0.5
e
1
1.5
a
0.5
b
0.5
c
0.5
2
a
0.25
b
0.25
c
0.25
d
0.25
e
0.25
f
0.25
2
a
0.50
b
0.50
c
0.50
d
0.50
Total Marks 10
Car Accident
Myopic
Study Survey
Name: Rahul Kaushal
Student ID: 101453613
List Price
$ 204,900 Question Part I
$ 210,000 Minimum Value
$ 204,900 $ 239,900 Maximum Value
$ 2,850,000 $ 269,000 Range
$ 2,645,100 $ 279,900 First Quartile
$ 825,500 $ 289,900 Second Quartile (Median)
$ 1,248,500 $ 297,900 Third Quartile
$ 1,379,975 $ 298,800 Inter Quartile Range
$ 554,475 $ 309,900 30 Percentile
$ 852,500.0 $ 319,900 Lower Limit of Inner Fence
$ (6,212.50)
$ 349,800 Upper Limit of Inner Fence
$ 2,211,687.50 $ 349,900 Lower Limit of Outer Fence
$ (837,925)
$ 359,900 Upper Limit of Outer Fence
$ 3,043,400 $ 389,900 Percentage of Prices less than $500,000
12.9%
$ 398,800 Number of Mild Outliers
0
$ 469,900 Number of Extreme Outliers
4
$ 578,000 Sample Mean
$ 1,149,661 $ 605,000 Sample Standard Deviation
$ 509,858.97 $ 615,000 $ 628,000 $ 632,000 $ 635,000 $ 640,000 $ 649,900 $ 702,500 $ 742,000 Answer B.
Intervals
$ 756,000 1
$ 757,000 2
$ 778,200 3
$ 821,000 4
$ 825,000 5
$ 827,000 6
$ 832,100 7
$ 839,000 $ 839,500 $ 841,000 Answer C.
$ 850,000 $ 855,000 $ 858,000 $ 859,000 $ 886,000 $ 890,000 $ 899,100 Refer to t
neighbor
a. Use E
b. Organ
the frequ
point. c. Draw d. Draw e. Draw
shapes.)
40
45
50
55
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$ 921,000 $ 925,000 $ 985,000 $ 992,300 $ 995,000 $ 1,025,000 $ 1,165,000 $ 1,198,000 $ 1,214,900 $ 1,219,877 $ 1,228,500 $ 1,229,900 $ 1,229,900 $ 1,234,900 $ 1,235,800 $ 1,244,900 $ 1,247,000 Answer D.
$ 1,248,000 $ 1,248,000 $ 1,249,000 $ 1,249,900 $ 1,249,900 $ 1,250,000 $ 1,254,800 $ 1,258,000 $ 1,259,900 $ 1,265,000 $ 1,270,000 $ 1,274,000 $ 1,274,900 $ 1,275,900 $ 1,279,888 $ 1,279,900 $ 1,288,000 $ 1,292,500 $ 1,299,900 $ 1,300,000 $ 1,319,000 $ 1,319,900 $ 1,328,000 $ 1,329,000 Answer E.
$ 1,338,000 $ 1,349,500 $ 1,358,000 $ 1,359,800 This chart isn
Editing this s
0
5
10
15
20
25
30
35
Frequency
0
10
20
30
40
50
60
Frequency
$ 1,359,900 $ 1,365,000 $ 1,372,000 $ 1,378,877 $ 1,379,900 $ 1,380,000 $ 1,389,900 $ 1,391,000 $ 1,410,000 $ 1,450,000 $ 1,479,900 $ 1,485,000 $ 1,496,000 $ 1,496,000 $ 1,499,000 $ 1,565,900 $ 1,568,000 $ 1,598,000 $ 1,609,000 $ 1,621,500 $ 1,625,000 $ 1,635,000 $ 1,725,800 $ 1,792,100 $ 1,807,000 $ 1,835,000 $ 1,836,000 $ 1,850,000 $ 1,980,000 $ 2,120,000 $ 2,150,000 $ 2,164,900 $ 2,250,000 $ 2,253,000 $ 2,350,000 $ 2,850,000
Frequency
Relative Frequency
0 - $300,000
8
0.067
8
$300,001 - $600,000
9
0.075
17
$600,001 - $900,000
26
0.217
43
$900,001 - $1,200,000
8
0.067
51
1,200,001 - $1,500,00
52
0.433
103
1,500,001 - $1,800,00
9
0.075
112
>= $1,800,001
8
0.067
120
120
1.000
Ontario Housing Prices
Cumulative Frequency
Housing Price Analysis
the Real Estate data provided in Column A of this sheet, which is a sample of the rhood in Ontario. Excel functions to provide your answers next to to each of the descriptive statistics
nize the data into a frequency distribution with seven class intervals (
Outliers shou
uency table and provide the following additional columns: relative frequency, cum
the histogram to show the frequency distribution.
the histogram to show the cumulative frequency distribution.
w box-plot based on the values in blue area. (You can draw it manually by in
)
52
Ontario Housing Price vs Frequency
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n't available in your version of Excel.
shape or saving this workbook into a different file format will permanently break the chart.
300000
600000
900000
1200000
1500000
8
9
26
8
Housing Price 300000
600000
900000
1200000
1500000
180
6.67%
14.17%
35.83%
42.50%
85.83%
93
Ontario Housing Price vs Frequency/Cumulative Freque
Frequency
Cumulative %
Housing Price
NOTES*
There are total of 4 Outliers:
$2,250,000 $2,253,000 $2,350,000 $2,850,000
Cumulative %
Midpoint
6.67%
150000
14.17%
450000.5
35.83%
750000.5
42.50%
1050000.5
85.83%
1350000.5
93.33%
1650000.5
100.00%
1982450.50
housing prices from a s listed in the blue section.
uld not be included
). Expand mulative frequency and mid nserting the appropriate
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Bin
Frequency
Bounds
300000
8
300000
600000
9
600000
900000
26
900000
1200000
8
1200000
1500000
52
1500000
1800000
9
1800000
>1800000
8
>1800000
Bin
Frequency
Cumulative %
300000
8
6.67%
600000
9
14.17%
900000
26
35.83%
1200000
8
42.50%
1500000
52
85.83%
1800000
9
93.33%
>1800000
8
100.00%
1800000
>1800000
9
8
00000
>1800000
0.00%
20.00%
40.00%
60.00%
80.00%
100.00%
120.00%
3.33%
100.00%
ency
The U.S. National Highway Traffic Safety Administration gathers data concerning
The following probabilities were determined from the 1998 annual study (BAC is 2000,
Table 1042.)
P
(
BAC
= 0 | Crash with fatality) = .616
P
(
BAC
is between .01 and .09 | Crash with fatality) = .300
P
(
BAC
is greater than .09 | Crash with fatality) = .084
Over a certain stretch of highway during a 1-year period, suppose the probability
has been estimated that 12% of the drivers on this highway drive while their BAC
one fatality if a driver drives while legally intoxicated (BAC greater than .09).
a. State the Original Information b. Describe the Question c. Apply the formula and determine the probability of a crach with at least one fa
Answer:
a. State the Original Information
The given probabilities from the 1998 annual study are:
1. P(BAC = 0 | Crash with fatality) = .616
- This is the probability that a driver had a blood-alcohol content (BAC) of 0 giv
2. P(BAC is between .01 and .09 | Crash with fatality) = .300
- This is the probability that a driver had a BAC between 0.01 and 0.09 given th
3. P(BAC is greater than .09 | Crash with fatality) = .084
- This is the probability that a driver had a BAC greater than 0.09 given that the
Other information provided:
4. The probability of being involved in a crash that results in at least one fatality i.e, P(Crash with fatality) = 0.01.
5. The probability that a driver drives whilwe their BAC is greater than 0.09 is 12
i.e, P(BAC>0.09) = 0.12 b. Describe the Question
The question is asking for the probability of a car crash with at least one fatality i.e, P(Crash with fatality | BAC > .09) = ?
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c. Apply the formula and determine the probability of a crash with at le
> .09)
Using the Bayes' theorem,
P(A | B) = [P(B | A) * P(A)] / P(B)
Let:
- A represent the event "Crash with fatality" and
- B represent the event "BAC > .09"
From the given information, we know that:
- P(B | A) = P(BAC is greater than .09 | Crash with fatality) = 0.084 - P(A) = P(Crash with fatality) = 0.01 - P(B) = P(BAC>0.09) = 0.12 Substitute these values into the Bayes' theorem:
P(A | B) = [P(B | A) * P(A)] / P(B)
= [0.084 * 0.01] / 0.12
= 0.00084 / 0.12
= 0.007
Therefore, P(Crash with fatality | BAC >0.09) = 0.007 = 0.7% So, the probability of a crash with at least one fatality if a driver drives while lega
g the causes of highway crashes where at least one fatality has occurred. blood-alcohol content).
(Source: Statistical Abstract of the United States, y of being involved in a crash that results in at least one fatality is .01. It C is greater than .09. Determine the probability of a crash with at least atality if a driver drives while legally intoxicated (BAC > .09)
ven that there was a crash with at least one fatality.
hat there was a crash with at least one fatality.
ere was a crash with at least one fatality.
over a certain stretch of highway during a 1-year period is 0.01.
2%.
if a driver drives while legally intoxicated (BAC greater than 0.09).
P(A | B)
east one fatality if a driver drives while legally intoxicated (BAC ally intoxicated (BAC > .09) is 0.007 or 0.7%.
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P(A) =
0.01
P(B) =
0.12
P(B | A) =
0.084
) = [P(B | A) * P(A)] / P(B)
P(B | A) =
0.007
Used Formula
p =
0.4
q = 0.6
n =
25
Mean (µ) = n*p 10
Standard Deviation = 2.449
a. What is the probability that 10 of them will become myopic befo
P(10) = P(x=10) = 0.161158
By using the excel formula =BINOM.DIST
b. What is the probability that fewer than 5 of them will become my
P(fewer than 5) = P(x<4) =
0.009471
By using the excel formula =BINOM.DIST
c. What is the probability that more than 15 of them will become m
P(more than 15) = P(x>15) = 1 - P(x<=15)
P(x<=15) = 0.9868309
1 - P(x<=15) = 0.013169
By using the excel formula =BINOM.DIST
d. What is the probability that at least 3 of them will become myopic P(at least 3) = P(x>=3) = 1 - P(x<=2)
Probability of Being Myopic
Researchers at the University of Pennsylvania School of Medicine theorized th
old who sleep in rooms with the light on have a 40% probability of becoming Suppose that researchers found 25 children who slept with the light on before
a. What is the probability that 10 of them will become myopic before age 16?
b. What is the probability that fewer than 5 of them will become myopic befor
c. What is the probability that more than 15 of them will become myopic befo
d. What is the probability that at least 3 of them will become myopic before a
e. What is the probability that at most 20 of them will become myopic before f. How many children will be expected to become myopic before age 16?
P(x<=2) =
0.000429
By using the excel formula =BINOM.DIST
1 - P(x<=2) =
0.999571
e. What is the probability that at most 20 of them will become myopic
P(at most 20) = P(x<=20) P(x<=20) = 0.999992
By using the excel formula =BINOM.DIST
f. How many children will be expected to become myopic before age
E(x) = n*p
E(x) = 25*0.4
E(x) = 10
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ore age 16?
yopic before age 16?
myopic before age 16?
before age 16?
hat children under 2 years myopic by age 16. e they were 2.
?
re age 16?
ore age 16?
age 16?
age 16?
Answer:
Given Information:
-Probability of becoming myopic by age 16 = 40%
So, p (probablity of sucess) = 0.4 and, q (probability -Number of children i.e, n = 25
Using Binomial model for Bernoulli's trials:
P(x) = (n!/(x!(n-x)!)) * p^x *(1-p)^n-x
Where:
- P(X=x) is the probability of k successes in n trials
- p is the probability of success on an individual trial
- n is the number of trials
-Mean = µ = n*p
Standard Deviation, S.D(σ) = √(n*p*q)
a. What is the probability that 10 of them will b
16?
P(10) = P(x=10) = 0.161158 b. What is the probability that fewer than 5 of before age 16?
P(fewer than 5) = P(x<4) = 0.009471
c. What is the probability that more than 15 of
before age 16?
P(more than 15) = P(x>15) = 1 - P(x<=15) P(x<=15) = 0.986831 1 - P(x<=15) = 0.013169 Therfore, P(more than 15) = P(x>15) = 1 - P(x<=15)
d. What is the probability that at least 3 of the
before age 16?
P(at least 3) = P(x>=3) = 1 - P(x<=2) P(x<=2) = 0.000429 1 - P(x<=2) = 0.999571 Therfore, P(at least 3) = P(x>=3) = 1 - P(x<=2) = 0
e. What is the probability that at most 20 of th
before age 16?
c before age 16?
e 16?
P(at most 20) = P(x<=20) P(x<=20) = 0.999992 Therefore, P(at most 20) = 0.999992
f. How many children will be expected to becom
Expected number of sucesses is the same as Mean
E(x) =Mean (µ) = n*p E(x) = 25*0.4 E(x) = 10 Therefore, expected children to become myopic befo
of failure = 1 - p) = 0.6
become myopic before age them will become myopic f them will become myopic ) = 0.013169
em will become myopic 0.999571
hem will become myopic
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me myopic before age 16?
ore age 16 = 10.
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Mean (µ) = 7.5
Standard Deviation (σ) = 2.1
Z = (X - μ) / σ
a. What proportion of A students study for more than 10 hours per week?
X = 10
Z = 1.19048
X>10) = P(Z > 1.190476) = 1 - P(Z < 1.190476
P(Z < 1.190476) = 0.88307
By using the excel formula =NORM.S.DIST
- P(Z < 1.190476) =
0.11693
P(Z > 1.190476) = 0.11693
P(X>10) =0.11693 or 11.69%
b. Find the probability that an A student spends between 7 and 9 hours s
X1 = 7
X2 = 9
Z1 = -0.2381
Z2 = 0.71429
P(Z1 < -0.2831) = 0.4059
(Z2 < 0.714286) =
0.76247
By using the excel formula
<X<9) = P(-0.2831< Z < 0.714286) = P(Z<-0.2831) - P(Z<0.7142
Z<-0.2831) - P(Z<0.714286) 0.35657
P(7<X<9) 0.35657
c. What proportion of A students spend fewer than 3 hours studying?
X = 3
Z = -2.1429
P(X<3) = P(Z<3)
P(Z < 3) =0.01606
By using the excel formula =NORM.S.DIST
d. What is the amount of time below which only 5% of all A students spe
Study Statistics
The amount of time devoted to studying statistics each week by students who ach
of A in the course is a normally distributed random variable with a mean of 7.5 hou
standard deviation of 2.1 hours.
a. What proportion of A students study for more than 10 hours per week?
b. Find the probability that an A student spends between 7 and 9 hours studying.
c. What proportion of A students spend fewer than 3 hours studying?
d. What is the amount of time below which only 5% of all A students spend studyin
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P = 5%
Z =
-1.6449
By using the excel formula =NORM.S.INV
Z = (X - μ) / σ
X = Zσ + μ
i.e
4.04581
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?
studying.
a =NORM.S.DIST
end studying?
hieve a grade urs and a ng?
Answer:
Given Information:
-Normally distributed random variables
-Mean (µ) = 7.5 hours
-Standard Deviation (σ) = 2.1 hours
a. What proportion of A students study for more than 10 First calculate the z-score for 10 hours and then subtract that pro
1.
The formula for z-score is:
Z = (X - μ) / σ
X=10
Z = (10 - 7.5) / 2.1 Z= 1.190476 P(X>10) = P(Z > 1.190476) = 1 - P(Z < 1.190476) P(Z < 1.190476) = 0.88307 1 - P(Z < 1.190476) = 0.11693 P(Z > 1.190476) = 0.11693
P(X>10) = 0.11693, i.e = 11.69%
Therefore, the probability of grade A students that study more th
0.11693 or 11.69% approximately.
b. Find the probability that an A student spends between
studying.
First calculate the z-scores for 7 and 9 hours, and then find the p
between these two z-scores.
Let,
X1 = 7, X2 = 9
Z1 = -0.2381 , Z2 = 0.714286 The proportion of values between Z1 and Z2 is the difference of proportions.
P(7<X<9) = P(-0.2831< Z < 0.714286) = P(Z<-0.2831) - P(Z<0.
P(Z<-0.2831) - P(Z<0.714286) = 0.356571 P(7<X<9) = 0.356571, i.e = 35.66%
Therefore, the probability that a grade A student spends between
studying is 0.3566, or 35.66% approximately.
c. What proportion of A students spend fewer than 3 hour
X = 3 Z = -2.14286 P(X<3) = P(Z<3) = 0.016062, i.e 1.61%
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P(X<3) = P(Z<3) = 0.016062, i.e 1.61%
Therefore, the probability of A students that spend less than 3 ho
0.0161, or 1.61% approximately.
d. What is the amount of time below which only 5% of all studying?
First find the z-score that corresponds to the lower 5% , and then
score to time in hours.
Probability = 5%
Z = -1.64485
Using the Z score formula: Z = (X - μ) / σ It can be said that, X =
X = 4.045807
Therefore, rhe amount of time below which only 5% of all A stude
hours approximately.
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hours per week?
oportion found from han 10 hours is n 7 and 9 hours proportion of values these two .714286) n 7 and 9 hours rs studying?
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ours studying is A students spend n convert this z-
= Zσ + μ
ents study is 4.05
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