Ms-8

IGNOU MBA MS-08 Solved Assignment 2011

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Course Code : MS 08

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Course Title : Quantitative Analysis for Managerial Applications

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Assignment No. : 08/TMA-1/SEM-I/2011

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Coverage : All Blocks

Note: Answer all the questions and send them to the Coordinator of the Study Centre you are attached with.

1. Calculate the mean, median and mode from the following data relating to production of a steel mill for 60 days

Production (in tons per day) | 21-22 | 23-24 | 25-26 | 27-28 | 29-30 |

Number of days | 7 | 13 | 22 | 10 | 8 |

Solution :

class | No. of days(f) | Mid value (x) | d=(x-A)/h | d2 | fd | fd2 | c.f. |

20.5-22.5 | 7 | 21.5 | -2 | 4 | -14 | 28 | 7 |

22.5-24.5 | 13 | 23.5 | -1 | 1 | -13 | 13 | 20 |

24.5-26.5 | 22 | 25.5 = A | 0 | 0 | 0 | 0 | 42 |

26.5-28.5 | 10 | 27.5 | 1 | 1 | 10 | 10 | 52 |

28.5-30.5 | 8 | 29.5 | 2 | 4 | 16 | 32 | 60 |

| ∑f=60 | | | | ∑fd= -1 | ∑fd2= 83 | |

a) Mean = A + [ (∑fd/∑f) × h]

Where A = assumed mean

h = class size

= 25.5 + [(-1/60) × 2]

= 25.467 (approx)

b) Median (Q2) = l1 + [{(N/2- p.c.f)/f} × l2 – l1

Median class = 24.5 – 26.5

Median = 24.5 + [{(60/2 – 20) /22} × 2]

= 25.409 (approx.)

c) Mode = l + [ { (f-f1)/(2f – f2 – f1) }× h]

Modal class = 24.5-26.5

Mode = 24.5 + [{ (22-13)/(2×22-13-10) } ×2]

= 25.357 (approx.)

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2. A restaurant is experiencing discontentment among its customers. It analyses that there are three factors responsible viz. food quality, service quality and interior décor. By conducting an analysis, it assesses the probabilities of discontentment with the three factors as 0.40, 0.35 and 0.25 respectively. By conducting a survey among the customers, it also evaluated the probabilities of a...