SOLUTION: 1)On average, 20 percent of the emergency room patients at Bedford hospital lack health insurance. In a random sample of 8 patients what is the probability that 2 will be uninsure

Question 628151: 1)On average, 20 percent of the emergency room patients at Bedford hospital lack health insurance. In a random sample of 8 patients what is the probability that 2 will be uninsured?

2)On average, 20 percent of the emergency room patients at Bedford hospital lack health insurance. In a random sample of 8 patients what is the probability that 2 will be insured?

3On average, 20 percent of the emergency room patients at Bedford hospital lack health insurance. In a random sample of 8 patients what is the probability that no more than 1 patient is uninsured?

4) Consumers report on average 1.7 problems per vehicle with a popular make and model of a 2006 car. we randoly select a vehicle of the same make, model and year. what is the probability that it will have more than 3 problems

5. At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.5 per day on a typical wednesday. What is the probability that less than 3 cancellations will occur on 2 wednesdays?

6 Thje pediatrics unit at Bedford hospital has 24 beds. The number of patients needing a bed at any point in time is normally distributed with a mean of 19.2 and standard deviation of 2.5. What is the probability that the number of patients needing a bed will exceed the pediatric unit bed capacity
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

 

Hi,

 1- 3 Bionimal Distribution Problems

Note: The probability of x successes in n trials is: 

P = nCx*  where p and q are the probabilities of success and failure respectively. 

In this case p = .20(uninsured & q = .80(insured)

nCx = 

1. P(x= 2) = 8C2 (.20)^2(.80)^6 = .2936

2. P(x=6) = 8C6(.20)^6(.80)^2   = .0011

3. P(x<1) = P(x=0) + P(x=1) = .1678 + .3355 = .5033

Using Normal distribution for #4 with SD = 1, therefore z = 1.3 and P(>3) = 1 - P(z=1.3) find 1 - NORMSDIST(1.3)

Using Poison Distribution for #5, Use stattrek.com Poison Distribution Calculator (WED and WED <3) = 

Using Normal distribution for #4 with mean of 19.2 and standard deviation of 2.5

P(x>24) = 1 - P(z = (24-19.2)/2.5) find 1 - NORMSDIST(1.92) 

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