Managerial Statistics

1. PGPMX 2015-17
Managerial Statistics
Assignment
-Sanmeet Dhokay
-PGPMX 025

2. Case: U. S. Public Health
Assume that p is the probability thatan individual is infected
n is the group size
N is the total no of men
UsingBinomial Distribution or otherwise,suggest the following:
1. Probability that a group of n individuals is not infected
So as given,
p = probability thatan individual isinfected
Then
(1 – p) = probability thata selected individual isnotinfected
Also
n = group size
Hence, probability thatgroup of individualsisnotinfected will be
(1 – p) * (1 - p) * (1 - p) ….. n times = (1 - p) n
Answer:- (1 - p) n
2. If X is the no of infected groups, what is the expected no. of infected groups?
X = No of infected groups
Let E(X) = Expected number of infected groups
From Question 1:-
Probability thatgroup of n individualsisnotinfected = (1 - p) n
Now ,
Probability thatat leastone member in group of n individualsisinfected (Pi)= 1 - (1 - p)n

3. N = total no of members
n = group size
Number of members in one group = N/n
Expected number of infected groups = Pi * (N/n) = [1 - (1 - p)n ]* (N/n)
3. Would the proposed technique require lesser no. of chemical analyses than those required by the
existing procedure?
From above,
Expected number of infected groups = Pi * (N/n)
Expected number of chemical analyses required E(C)
= Number of groups + Number of times the chemical analyses needs to be repeated insidean
infected group
E(C) = N/n + Pi * (N/n) * n
= N/n + [1 - (1 - p)n] *(N/n) * n ……………………………. (a)
So proposed technique require lesser no. of chemical analyses than thoserequired by the existing
procedure , as dividingthe N men into groups of sizen would reduce the number of chemical analyses
required as compared to the existingprocedure which would subjecteveryone to the test.
Also if the probability p that the person is infected is less,then the number of chemical analyses would be
reduced as per the formula derived(a) for the proposed technique.

4. Case: Speciality Toys:
Prepare a managerial reportthat addresses the followingissues and recommends an order quantity for the
Weather Teddy product.
You may assumethat the demand for the product follows normal distribution.
1. Compute the probability of a stock-out for the order quantities suggested by membersof the
management team.
Let Demand for the Weather Teddy product = D
Assumingthat D follows Normal Distribution,letμ be the Mean and σ be the Standard Deviation
As per given information in the problem , predicted expected demand is 20000 units with a 0.90
probability thatdemand would be between 10000 and 30000 units
So , μ = 20000
P(10000 < D < 30000) = 0.90
Usingformula , Z= (x- μ)/ σ
P(10000 - 20000)/ σ < (D – 20000)/ σ < (30000 – 20000)/ σ = 0.90
P(Z) = P[(1-0.90)/2] = P(0.1/2) = 0.05
From Table, Z=1.64
σ = (30000 - 20000)/1.64
Standard Deviation(σ) = 6097.560976
~ 6098
Let O be the order quantity
Probability of an order stock out will be when Demand D is greater than Order O.
P(D > O) = P(Z > (O-20000)/6098)
Order Quantity (O)
Z=(O-
20000)/6098 P(D > O) (Values from Normal DistTable)
15000 -0.819940964 (1-0.2061)=0.7939
18000 -0.327976386 (1-0.3707)=0.6293
24000 0.655952771 0.2546
28000 1.311905543 0.0951

5. 2. Compute the projected profit for the order quantities suggested by the management team under
three scenarios: worst case in which sales = 10,000 units, most likely case in which sales = 20,000
unit, and best case in which sales = 30,000 units.
Sellingpricefor season = $24
Cost per unit = $16
Profitper unit= $24 - $16 = $8
SellingPriceif inventory remains = $5
Loss per unit if inventory remains = $16 - $5 = $11
Profitcalculationsfor different order quantities and correspondingsales
Order
Quantities
Sale = 10000 Units Sale = 20000 Units Sale = 30000 Units
15000 8*10000-11*5000 = $25000 8*15000 = $120000 8*15000 = $120000
18000 8*10000-11*8000 = -$8000 8*18000 = $144000 8*18000 = $144000
24000
8*10000-11*14000 = -
$74000
8*20000 -11*4000 = $116000 8*24000 = $192000
28000
8*10000-11*18000=-
$118000
8*20000-11*8000 = $72000 8*28000 = $224000
3. One of Specialty's managers felt that the profit potential was so great that the order quantity
should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What
quantity would be ordered under this policy, and what is the projected profit under the three sales
scenarios?
Calculation of the order quantity to have 30% chanceof stockout
30% Probability of Stockout P(Z) = 0.3
From Normal DistTable Z ~ 0.525
So usingformula Z= (X- μ)/ σ
X= (Z * σ) + μ
X=(0.525 * 6098) + 20000

6. X = 23201
Order quantity should be 23201 for 30% Probability of stockout
4. Provide your own recommendation for an order quantity and note the associated profit
projections. Provide a rationale for your recommendation.
Order
Quantities
Sale = 10000 Units Sale = 20000 Units Sale = 30000 Units
15000 8*10000-11*5000 = $25000 8*15000 = $120000 8*15000 = $120000
18000 8*10000-11*8000 = -$8000 8*18000 = $144000 8*18000 = $144000
24000
8*10000-11*14000 = -
$74000
8*20000 -11*4000 = $116000 8*24000 = $192000
28000
8*10000-11*18000=-
$118000
8*20000-11*8000 = $72000 8*28000 = $224000
23201
8*10000-11*13201=-
$65211
8*20000-11*13201 = $14789 8*23201 = $185608
19000 8*10000-11*9000=-$19000 8*19000 = 152000 8*19000 = 152000
20000
8*10000-11*10000=-
$30000
8*20000 = 160000 8*20000 = 160000
21000
8*10000-11*11000=-
$41000
8*20000 - 11*1000 = 149000 8*21000 = 168000
Given that Probability is0.9 for sales between 10000 and 30000 based on the previous experience
and as per the senior sales forecaster.
Also,we have to consider the profitper unit of $11 and loss per unitof $8 if inventory remains.
Takingall this into consideration ,order quantity of 20000 looks to be the best as per the above table
and optimisingthe profit/loss
Calculations are present in attached sheet
Speciality Toys
Calculations.xlsx
Case: Lorex Pharmaceuticals:
1. What is the issue facing Carter Blakley?
Carter Blakley ,manager of quality assurancefor the manufacturingdivision of Lorex Pharmaceuticals,
is faced with the issueof determining the fill targetfor the Linatol bottles which were packed in
quantities of 10 ounces each.
2. What is wrong with one standard deviation rule? What other factors need to be considered?

7. One standard deviation ruleis to pick a fillingtargetwhich is one standard deviation abovethe
required amount. As the target amount can be towards both sides of the mean and because of certain
economic factors,itlead to some bottles getting under filled thereby cloggingthe buffer storage area
with rejected bottles and leadingto temporary stoppage of the entire fillingline.Economic factors
related to the production likeusageof older and slower product lines, semiautomatic filling
process(manual intervention required) should also beconsidered
3. What analysis must we do to incorporate all these factors (that you mentioned in the previous
question) correctly?
We have to find the exact target amount in ounces which will reduce/minimizethe number of
rejected bottles and maximiseprofit.The target amount also can’tbe excess as it will reducethe
number of production bottles.
Analysis can also bedone regardingany implementation of new technology to reduce the standard
deviation of 0.20 ounces
Handlingof rejected bottle process can be improved to lower the losses as thereis 20%difference in
the cost of good vs rejected bottle
Cost associated with 1 ounce :-
15.5 $ per 10 ounces(12 bottles of 10 ounces each in each caseand $186 per case)
Production costs :-
Category Cost
Cost of materials per bottle $1.10 per bottle
Production lineoperated by 2 employees $12.80 per hour per
employee
=$25.6 per hour
Overhead rate for production line $89.50 per hour
Under filled bottleattendants $8.50 per hour
500 cases or (500*12= 6000 bottles) were produced in eight hour shift
So per hour production = 6000/8 = 750 bottles
Total Cost per hour = 25.6 + 89.50 + 8.50 = $123.6
Total Cost for producingone bottle = (750/123.6) + 1.10 = $7.17

8. As per data given in Exhibit 1,
Total Liters of Linatol = 5000
Total Ounces of Linatol = 169088
Total Costs associated = $ 90888
Total cost per ounce = 169088/ 90888 = $0.5375 ~ $0.54
As per exhibit 2,
X would always be10 as the bottles should have 10 ounces
We have to estimate μ which would be our target
Sample calculation for μ= 10 , X = 10, σ = 0.16
Usingformula , Z= (x- μ)/ σ
Z = 10 – 10/0.16
= 0
So referring the normal distribution table,P(Z) = 0.5.
We can also find the P(Z) usingexcel formula = NORMDIST(10,10,0.16,TRUE)
μ = 10.20
σ = 0.16
Now we can vary x and find Z and P(Z) , we can start from the value10
Calculations are present in the attached sheet
Lorex Case
Calculations.xlsx
4. Lorex is really sold in 6-packs. What is the forecast of the total number of ounces in a 6-pack?
Total Ounces = 10.2* 6 = 61.2
σ = 0.2
Standard Deviation for 6 bottles would be = 0.2/Sq Root(6) = 0.2/2.449 = 0.082 ounces

9. Forecastfor the total number of ounces in 6 Bottle pack would be = 61.2 – 2*(0.082) to 61.2 +
2*(0.082)
= 61.036 to 61.364 ounces
5. Assume that Lorex is not able to rework underfilled bottles. To what target amount should the
bottles be filled in order to maximize the profit per case of Linatol?
Target Amount to which the bottles be filled would be:
It can be 10.38 to 10.40 ounces to maximisethe profit
Calculations are present in the attached sheet
Lorex Case
Calculations.xlsx