Question

1) 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?

2)On an immediate memory test, 10-year-old children can correctly recall an average of μ=7 digits. The distribution of recall scores is normal with σ=2. a. What is the probability of randomly selecting a child with a recall score less than6? b. What is the probability of randomly selecting a sample of n=4 children whose average recall score is less than 6?

Answer 1

Solution :

Given that ,

1) mean = = 12.0

standard deviation = = 1.0

a) n = 100

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

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

=1- P(z < 1.00 )

Using z table,

= 1 - 0.8413

= 0.1587

= 0.1587 * 100 = 15.87 = 16

b) x = 11.3

Using z-score formula,

z = x - /   

z = 11.3 - 12.0 / 1.0

z = -0.70

= P(z < -0.70)

Using z table,

= 0.2420

the percentage is = 24.20%

The percentile is 24 th

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

= P( -0.50 < z < 1.00)

= P(z < 1.00) - P(z < -0.50)

Using z table,

= 0.8413 - 0.3085

= 0.5328

The percentage is = 53.28%

2) mean = = 7

standard deviation = = 2

a) P(x < 6) = P[(x - ) / < (6 - 7 ) / 2]

= P(z < -0.50 )

Using z table,

= 0.3085

b) n = 4

=   = 7

= / n = 2 / 4 = 1

P( < 6) = P(( - ) / < (6 - 7) / 1)

= P(z < -1.00)

Using z table

= 0.1587