Question

In a population survey of patients in a rehabilitation hospital, the mean length of stay in the hospital was 12.0 weeks with a standard deviation equal to 1.0 week. The population distribution was normal.

a. Out of 100 patients how many would you expect to stay longer than 13 weeks?

b. What is the percentile rank of a stay of 11.3 weeks?

c. What percentage of patients would you expect to stay between 11.5 weeks and 13.0 weeks?

Answer 1

Solution :

Given that ,

mean = = 12.0

standard deviation = = 1.0

P(x >13 ) = 1 - P(x<13 )

= 1 - P[(x -) / < (13 -12.0) /1.0 ]

= 1 - P(z <1 )

Using z table

= 1 - 0.8413

= 0.1587

answer =15.87

(B)

P(X< 11.3) = P[(X- ) / < (11.3 -12.0) /1.0 ]

= P(z <-0.7 )

Using z table

=0.242

percentile rank =24.2%

(c)

P(11.5< x <13.0 ) = P[(11.5 -12.0) / 1.0< (x - ) / < (13.0 -12.0) / 1.0)]

= P(-0.5 < Z <1 )

= P(Z <1 ) - P(Z <-0.5 )

Using z table   

= 0.8413 -0.3085

= 0.5328

percent=53.28%