Statistics Group 15 Kilgore Manufacturing Inc
Introduction to Statistics
Assignment 1 – Kilgore Case Study
Group Members
Jayadeep P.
Janani Sekar
Karthik Shankar
Manya Arora
Mathura Vaidyanathan
Date of Submission – 4th August 2015
Executive Summary
Kilgore Manufacturing Inc. (KMI) has the opportunity to win a defence contract with Avionics by producing R-7 switches. KMI faces stiff competition from three competitors, all of whom will employ their resources in order to bid for extremely high production limits. The winning bidder must meet their minimum production guarantee failing which a hefty penalty of $5000 will be imposed on each day that the production falls short of the guaranteed minimum units.
Under its old production process KMI can afford to bid 612 units which will ensure it has a profit of $1000 per day. However, the bid may not be high enough to win over its cutthroat competitors and KMI introduced a new production process to improve its efficiency.
Under the new production process KMI has the same average but a new standard deviation which reduces its possible bid amount to 607 units. However, buried in the figures is the fact that in the second and third month, the production stabilises which potentially allows KMI to propose a higher production figure. This paper analyses both production methods for KMI and proposes the optimal number in order to win the bid with Avionics.
Analysis
Production Process I
Under the old production process, KMI had a mean ( of 635 units and a standard deviation ( of 40 units.
We need to find a value of y such that
$2413 - $5000 * P(Y < y) = $1000
Solving for this equation we get $5000 * P(Y < y) = 1413
=> P(Y < y) = 1413/5000
=> P(Y < y) = 0.2826
On the negative Z table, this value corresponds to -0.575
Given that = Z
Plugging in the values of we get = - 0.575
Hence we get = 612 µy = 635 σy = 40 0.2826
612 635
Assignment 1 – Kilgore Case Study
Group Members
Jayadeep P.
Janani Sekar
Karthik Shankar
Manya Arora
Mathura Vaidyanathan
Date of Submission – 4th August 2015
Executive Summary
Kilgore Manufacturing Inc. (KMI) has the opportunity to win a defence contract with Avionics by producing R-7 switches. KMI faces stiff competition from three competitors, all of whom will employ their resources in order to bid for extremely high production limits. The winning bidder must meet their minimum production guarantee failing which a hefty penalty of $5000 will be imposed on each day that the production falls short of the guaranteed minimum units.
Under its old production process KMI can afford to bid 612 units which will ensure it has a profit of $1000 per day. However, the bid may not be high enough to win over its cutthroat competitors and KMI introduced a new production process to improve its efficiency.
Under the new production process KMI has the same average but a new standard deviation which reduces its possible bid amount to 607 units. However, buried in the figures is the fact that in the second and third month, the production stabilises which potentially allows KMI to propose a higher production figure. This paper analyses both production methods for KMI and proposes the optimal number in order to win the bid with Avionics.
Analysis
Production Process I
Under the old production process, KMI had a mean ( of 635 units and a standard deviation ( of 40 units.
We need to find a value of y such that
$2413 - $5000 * P(Y < y) = $1000
Solving for this equation we get $5000 * P(Y < y) = 1413
=> P(Y < y) = 1413/5000
=> P(Y < y) = 0.2826
On the negative Z table, this value corresponds to -0.575
Given that = Z
Plugging in the values of we get = - 0.575
Hence we get = 612 µy = 635 σy = 40 0.2826
612 635