Answer:
a) the probability of a length of stay greater than 10 hours is 0.03144
b) the length of stay is exceeded by 25% of the visits is 6.5 hrs
c)
P( X < 0 ) = 0.0559
The probability of length of stay less than 0 hours is 0.0559, length of stay cannot be less than 0. When the normal model is used, we assume that the LOF is approximately normally distributed.
The LOF is better modeled by the normal distribution near the mean and worse in the tails.
The probabilities in the left tail for the vales of X < 0 can be neglected.
Explanation:
Given that;
mean μ = 4.6
standard deviation σ = 2.9
let x represent length of stay;
a. What is the probability of a length of stay greater than 10 hours?
p( x > 10 ) = p( x-μ/σ > 10-4.6 / 2.9 )
= p( Z > 1.86)
= P( Z < - 1.86 )
FROM THE Z SCORE TABLE;( Z < - 1.86 ) = 0.03144
p( x > 10 ) = 0.03144
Therefore, the probability of a length of stay greater than 10 hours is 0.03144
b) What length of stay is exceeded by 25% of the visits?
p( X > x ) = 0.25
so
p( X < x ) = 0.75
p( x-μ/σ < x-4.6 / 2.9 ) = 75
p( Z < x-4.6 / 2.9 ) = 0.75 ----- 1
also, from the standard normal table; P( Z < 0.67 ) = 0.75 ------ 11
from equation 1 and 11
x-4.6 / 2.9 = 0.67
x-4.6 = 2.9 × 0.67
x - 4.6 = 1.943
x = 4.6 + 1.943
x = 6.543 ≈ 6.5
Therefore, the length of stay is exceeded by 25% of the visits is 6.5 hrs
c) From the normally distributed model, what is the probability of a length of stay less than 0 hours?
P( X < 0 ) = p( x-μ/σ < 0-4.6 / 2.9 )
= P( Z - 1.59)
from table;( Z - 1.59) = 0.0559
P( X < 0 ) = 0.0559
The probability of length of stay less than 0 hours is 0.0559, length of stay cannot be less than 0. When the normal model is used, we assume that the LOF is approximately normally distributed.
The LOF is better modeled by the normal distribution near the mean and worse in the tails.
The probabilities in the left tail for the vales of X < 0 can be neglected.