Question

(3) Suppose the length of stay in a chronic disease hospital of a certain type of patient has the mean of 60 days with a standard deviation of 15 days. It is reasonably to assume an approximately normal distribution of the lengths of stay (a) If one patient is selected from this group at random, what the probability that this patient will have a length of stay between 50 and 90 days? at is selected from this group at random, what is the probability that this patient will have a length of stay more than 110 days? (c) (Extra) Suppose a patient needs to receive an extra care if 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 an extra care?

Answer 1

Answer

we have $\mu=60, \sigma = 15$ (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

$P(x_1<X<x_2) = P((x_1-\mu)/\sigma<z<(x_2-\mu)/\sigma)$

where x1 = 50, x2 = 90 and $\mu=60, \sigma = 15$

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 $P(X>x) = P(z>(x-\mu)/\sigma)$

setting x = 110, $\mu=60, \sigma = 15$

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 = $(\bar{x}-\mu)/\sigma$

where z= -1.64, $\mu=60, \sigma = 15$ 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