The income of a group of 10,000 persons was found to be normally distributed with mean Rs.750 per month and a standard deviation of Rs. 50, show that of this group about 95% has income exceeding Rs. 668 and only 5% had income exceeding Rs. 832. (area between 750 and 668 = 0.4495, area between 750 and 832 = 0.4495)

                                                                                                                                           

MBA

Quantitative Analysis for Managerial Applications

ASSIGNMENT

 

Course Code: MMPC-005

Assignment Code: MMPC-005/TMA/JULY/2022 

Coverage : All Blocks

1. The income of a group of 10,000 persons was found to be normally distributed with mean Rs.750 per month and a standard deviation of Rs. 50, show that of this group about 95% has income exceeding Rs. 668 and only 5% had income exceeding Rs. 832. (area between 750 and 668 = 0.4495, area between 750 and 832 = 0.4495).

The problem provides us with information about a normally distributed group of 10,000 persons with a mean income of Rs.750 per month and a standard deviation of Rs.50. The goal is to calculate the proportion of the group with income exceeding Rs. 668 and Rs. 832. 

To begin, we can use the formula for the z-score: z = (x - mu) / sigma where x is the income value, mu is the mean income, and sigma is the standard deviation. Using this formula, we can calculate the z-score for an income of Rs.668 as: z = (668 - 750) / 50 = -1.64 Similarly, we can calculate the z-score for an income of Rs.832 as: z = (832 - 750) / 50 = 1.64

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