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Karolina [17]
3 years ago
7

8 Hospital administrator Jake Rosen9 was recently con victed for fraud he committed against his employer, Cedar Hospital Systems

. Over a period of six years, he allegedly made payments to a dummy company for maintenance charges while simultaneously running a scheme with maintenance contractors where he either paid them for work never performed or overpaid them for work. The skyscraper where Jake worked was only 10 years old. Maintenance charges rose from $5.2 mil lion in 1994 to $16.4 million in 2000. It was worth noting that the judge on the case questioned whether Cedar Hospital Systems deserved less than full restitu tion for failing to notice the problem. However, it was determined that federal law on restitution does not al low such charges, so Jake Rosen will be making monthly payments towards the alleged $8 million he stole until he makes full restitution after leaving prison. He was able to quickly repay $3.2 million of the theft with assets recovered by the government, including two homes and a nice yacht. Not bad for a man who was supposed to be making $90,000 per year. 1. In what specific types of fraud was Jake Rosen engaged
Business
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Asset Misappropriation.

Explanation:

The type of fraud that Jake Rosen engaged in is called Asset Misappropriation.

Asset Misappropriation happens when a person diverts the assets of the company they work for or the client they represent, for their own personal use.

Jake Rosen diverted the funds of the hospital for his own personal use to enabled the purchase of two homes and a nice yacht amongst other things thereby making him guilty of asset misappropriation.

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Answer:

a) 15 to 35 approximately 95%

(b) 10 to 40 approximately almost all

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Explanation:

The data have a bell-shaped distribution which means the data is equally distributed on both sides of the mean.

We have the mean at 25 and a standard deviation of 5 which means that the interval is for each of the values of 5 .

The mean would be u and

The first value would be u ±σ = 25 ± 5= 20 and 30 (68 % )

The second value will be u ± 2σ= 25± 10 = 15 and 35 (95%)

The third value will be u ± 3σ= 25 ± 15 = 10 and 40 (99.7 % almost all)

In the figure below the light blue region gives u ±σ on both sides of the mean

, dark blue gives u ± 2σ values on both sides of the mean and grey gives

u ± 3 σ values on both sides of the mean.

It is obvious that 68 % of the data is contained in the u ±σ light blue region, 95 % of the data in the  u ± 2σ dark blue including light blue and 99.7 % in the u ± 3σ all colored regions.

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