Calculations and Statistical Graphs

Calculations and Statistical Graphs

TABLE OF CONTENTS
QUESTION 1...................................................................................................................................1
QUESTION 2...................................................................................................................................1
QUESTION 3...................................................................................................................................3
A) P(A)...................................................................................................................................3
B) P(A ǀ B)..............................................................................................................................4
C) Events for A and B for independent..................................................................................4
QUESTION 4...................................................................................................................................4
A) Value of randomly selected property is of $1 million.......................................................4
B) Probability of having sales less than $2 million................................................................4

QUESTION 1
Histogram
Row
Labels
Count of Lot
Size
460-529 9
530-599 11
600-669 18
670-739 10
740-809 12
Grand
Total 60
QUESTION 2
Computation of mean,1 median, standard deviation and range (in the case of individual data
series)
Lot Size
Mean 634.67
Median 640
Standard
Deviation 90.36
Range 330
Minimum 460
1

Maximum 790
Descriptive statistics (in the case of continuous data series)
Lot
Size
Frequency
(F)
X (mid
value) FX CF
460-
529 9 495 4450.5 9
530-
599 11 565 6209.5 20
600-
669 18 635 11421 38
670-
739 10 705 7045 48
740-
809 12 775 9294 60
60 38420
Calculation of mean:
Mean = ∑FX/∑F
= 38420 / 60
= 640.33
Computation of median
Median = ∑F/2
= 60/2
= 30
This figure lies in the class interval of 600-669
Median= L1+ [(N/2 – C)/F]* i (difference between the class interval)
M = 600 + [ (60/2 – 20) / 18] * 70
M = 600 + 38.89
= 638.89 or 639 approx.
Standard deviation
2

Lot
Size
Frequency
(F)
X (mid
value) FX CF X^2 FX^2
460-
529 9 495 4450.5 9 244530.3 2200772
530-
599 11 565 6209.5 20 318660.3 3505263
600-
669 18 635 11421 38 402590.3 7246625
670-
739 10 705 7045 48 496320.3 4963203
740-
809 12 775 9294 60 599850.3 7198203
38420 2061951 25114065
Variance (S^2) = ∑Fx2 – ((∑Fx)2 /n)/ n – 1
= 25114065 – (38420) ^2 /60 / (60-1)
= 512458.33 / 59
= 8685.73
Standard deviation = √ Variance
= √ 8685.73
= 93.20
Range
Highest value – lower value
809 – 460
= 349
QUESTION 3
A) P(A)
There is a number which assigned between 0 and 1 to an event. Therefore P(A)= 1 for
fireplace and P(B) = ½ for Baths.
P(A) =1/2
P(B) = 2
3

B) P(A ǀ B)
P(A ǀ B) = P(A . B)/P(B)
P(A ǀ B) = P( 1*1)/P(2)
P(A ǀ B) = ½ =0.5
C) Events for A and B for independent
P (A ǀ B) has denoted both variables as independent where event A occurred which does
not affect any changes in probability of B. Thus, as per considering the probability of having 2
bathrooms and a fireplace which insist that there will be no relation on each other for dependent
conditions. However, there have been two equivalent ways such as:
P (B ǀ A) = P(B)
P (A and B) = P (B ∩ A) = P(B) * P(A)
QUESTION 4
A) Value of randomly selected property is of $1 million
To analyses the probability of a randomly selected property can be analysed as
= X- μ/ 𝞼
= $650000- $1000000/ $S150000
= -2.33 or -2*1/3
However, as per the negative outcomes the probability of property to be valued is -2.33
or -2*1/3
B) Probability of having sales less than $2 million
= X- μ/ 𝞼/ √n
= $650000- $2000000/ $150000- √2000000
= -9
4