# Analysis of Flower Cost and Number of Flowers

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|
|
|
Amount
spent
Lower
B
Mid-
point x f Cf x.f μ x-μ (x-μ)^2 f. ((x-μ)^2)
0-10
-0.5
5 5 5 25 50.3
-
45.3
2052.0
9 10260.45
10-20
9.5
15 7 12 105 50.3
-
35.3
1246.0
9 8722.63
20-30
19.5
25 10 22 250 50.3
-
25.3 640.09 6400.9
30-40
29.5
35 12 34 420 50.3
-
15.3 234.09 2809.08
40-50 39.5 45 14 48 630 50.3 -5.3 28.09 393.26
50-60 49.5 55 15 63 825 50.3 4.7 22.09 331.35
60-70 59.5 65 14 77 910 50.3 14.7 216.09 3025.26
70-80 69.5 75 13 90 975 50.3 24.7 610.09 7931.17
80-90
79.5
85 6 96 510 50.3 34.7
1204.0
9 7224.54
90-100
89.5
95 4 100 380 50.3 44.7
1998.0
9 7992.36
∑f=100
∑x.f=50
30
∑x. f((x-
μ)^2)=55091
Question 1
Part a.
Mean,
μ= x . f / f
μ= 5030
100 =50.30
Median,
is the number in the middle
Therefore, medium 0.5(100) =50
Bur 50 falls within the 50-60 class and the value is 63.
Mode,
Occurs where the frequency is highest.
And from our table, we can see that the highest frequency is 15.
Now the modal class is 50-60
In conclusion, the average cost spent on the flowers is \$ 50.30, median
number of flower ranges is 63 and the modal money spent is between \$50 to
\$60.
From the data above, Elena’s sales and expenditure are normally
distributed hence she experiences no lost at the moment.
Part b
Standard deviation
s= ¿
s= (55091
99 )
s=23.5897
Part c
We will use the formula,
Qk=LB¿]
Where Qk is the quartile you are computing
Cfk=the cumulative frequency before the kth number.
fQk=frequency where the Qk falls.
1st Quartile.
Q1=25/100*100=25
Given N=100, LB=29.5, Cfb=22, fQ1=12, i=10
Q1=29.5+[(25-35)/12]*10=29.5+2.5=32.0
Therefore, the Q1= 32.0
3rd Quartile.
Q3=75/100*100=75
Given N=100, LB=59.5, Cfb=63 , fQ3=14, i=10
Q3=59.5+ [(75-63)/14)*10=59.5+8.57=68.07
Therefore, Q3=68.07
The lower quartile shows the least number of customers Elena can
serve in a day while the upper quartile shows the maximum number of
customers that Elena serves in a busy day.
Question 2
Loan Source Lss than \$ 20000 \$20000-\$60000
More than
\$60,000 Total
Building Society 40 65 45 150

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