In: Statistics and Probability
The table shows the daily sales (in $1000) of shopping
mall for some randomly selected
days
Sales 1.1-1.5 1.6-2.0 2.1-2.5 2.6-3.0 3.1-3.5 3.6-4.0 4.1-4.5
Days 18 27 31 40 56 55 23
Use it to answer questions 13 and 14.
13. What is the approximate value for the modal daily sales?
A. $3,129.41 B. $2,629.41 C. $3,079.41 D. $3,123.53
14. The approximate median daily sales is …
A. $3,130.36 B. $2,680.36 C. $3,180.36 D. $2,664
Classes |
Frequency (f) |
1.1-1.5 |
18 |
1.6-2.0 |
27 |
2.1-2.5 |
31 |
2.6-3.0 |
40 |
3.1-3.5 |
56 |
3.6-4.0 |
55 |
4.1-4.5 |
23 |
First we will convert our inclusive frequency distribution to exclusive
Class |
Frequency |
1.05 - 1.55 |
18 |
1.55 - 2.05 |
27 |
2.05 - 2.55 |
31 |
2.55 - 3.05 |
40 |
3.05 - 3.55 |
56 |
3.55 - 4.05 |
55 |
4.05 - 4.55 |
23 |
Solution:
Class |
Frequency
(f) |
cf |
1.05-1.55 |
18 |
18 |
1.55-2.05 |
27 |
45 |
2.05-2.55 |
31 |
76 |
2.55-3.05 |
40 |
116 |
3.05-3.55 |
56 |
172 |
3.55-4.05 |
55 |
227 |
4.05-4.55 |
23 |
250 |
--- |
--- |
--- |
-- |
n=250 |
-- |
14
To find Median Class
= value of (n / 2)th observation
= value of (250 / 2)th observation
= value of 125th observation
From the column of cumulative frequency cf,
we find that the 125th observation lies in the class
3.05-3.55.
∴ The median class is 3.05-3.55.
Now,
∴L=lower boundary point of median class =3.05
∴n=Total frequency =250
∴cf=Cumulative frequency of the class preceding the median
class =116
∴f=Frequency of the median class =56
∴c=class length of median class =0.5
Median
M=L+n2-cff⋅c
=3.05+125-11656⋅0.5
=3.05+956⋅0.5
=3.05+0.08036
=3.13036
hence median sales = 3130.36
13.
Similarly modal daily sales =
we can estimate the Mode using the following formula:
Estimated Mode =
L + fm − fm-1(fm − fm-1) + (fm − fm+1) × w
where:
now
We can easily find the modal group (the group with the highest frequency), which is 3.05 -3.55
L = 3.05
fm-1 =40
fm =56
fm+1 = 55
w = 0.5
mode =
mode = 3.05 + 0.4705 = 3.5205
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