Question

The average length of stay in a chronic disease hospital for a certain type of patient is 60 days with a standard deviation of 15 days. If 36 of these patients are randomly selected, find the probability that the average length of stay of these 36 patients is less than 58 days.

(a) .1519 (b) .3176 (c) .4483 (d) .2119 (e) .6124

Answer 1

Solution :

Given that ,

mean = = 60

standard deviation =15

n=36

=average length of stay   in a chronic disease hospital

~Normal(,)

==60

= /sqrt(x)=15/6=2.5

#the probability that the average length of stay of these 36 patients is less than 58 days.

P(<58)=P((-)/<(58-60)/2.5)

=P(Z<-0.8)

=0.2119

#Value of z is obtain from standard

P(<58) =0.2119

#Option D is correct