Assume that a specific emergency department has a length of stay with behavior n

Question

Assume that a specific emergency department has a length of stay with behavior normally distributed; mean of 5 hours and a standard deviation of 3 hours. Use standard normal random variables to resolve these probabilities. What is the probability of a length of stay greater than 12 hours? What length of stay is exceeded by 20% of the visits? From the normally distributed model, what is the probability of a length of stay less than 0 hours? Comment on the normally distributed assumption in this example.

Explanation / Answer

Here we are given that Mean=5 hours and Standard deviation=sd =3 hours

a.Let X be the lenght of stay.

p(x>12)=p(X-MEAN/SD > 12-5/3)

=P(Z>2.33)

=1- P(Z<2.33)

=1-0.9902

=0.00982

b.We need to compute value of x such that P(X>x)=0.20

P(X-mean/SD>x-5/3)=0.20

P(Z>z)=0.20

P(Z<z)=0.80

From the standard normal table we get z for 0.80 probability is

z=0.842

x-mean/SD=0.842

x=5+0.842*3

=7.526

(c)P(X<0) = P(Z<(0-5)/3)

=P(Z<-1.66) =0.047 (from standard normal table)

The normally distributed assumption is satisfied

Hope this is helpful you. Thanks and good luck :)

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