Answer
we have (mean and standard deviation given in the question)
(A) we have to find the probability of a length stay between 50 and 90 days
using the formula
where x1 = 50, x2 = 90 and
setting the values, we get
on solving, we get
we can write it as
using the identity
So, we get
now using z distribution table to get the required p values, we get
So, the required probability is 0.7258
(b) We have to find the probability that the length stay is more than 110 days
i.e. we have to find the value of
using the formula
setting x = 110,
we get
on solving, we get
using the identity
we can write it as
using the z distribution, we get
So, the required probability
Required probability is 0.0004
(C) z score corresponding to the lower 5% of the data is -1.64 using z table
We have the formula for the z
z =
where z= -1.64, and we have to find the value of x(bar)
we get
-1.64 = (x-60)/15
multiplying both sides by 15, we get
-1.64*15 = x- 60........................(15/15 = 1 on the right hand side)
-24.6 = x- 60, adding 60 on each side, we get
60-24.6 = x ........................( -60+60 = 0 on right hand side)
we get
x = 35.4
So, the cutoff length of stay is 35.4 days
(3) Suppose the length of stay in a chronic disease hospital of a certain type of...
(a) ne pat h (b) If one patient is selected from patient e d from this up have a lenchotay more than 110 days that he (c) (Extra) Suppose a patient needs to receive an extra care if he/she stays in the hospital beyond a certain period and the hospital only want to have at most 5% of the patient to receive such an extra care. What is the cut-off length of stay for the chosen patient to receive such...
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