The length of stay at a specific emergency department in Phoenix, Arizona, in 2009 had a mean of 4.6 hours with a standard deviation of 2.9. Assume that the length of stay is normally distributed. a. What is the probability of a length of stay greater than 10 hours?
b. What length of stay is exceeded by 25% of the visits?
c. From the normally distributed model, what is the probability of a length of stay less than 0 hours? Comment on the normally distributed assumption in this example.

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.

Step-by-step 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.  

answer: i wpould go with c

step-by-step explanation:

answer: input the dimensions of the room to be painted. click the button to calculate the amount of paint and man-hours needed for this job.

interior paint coverage depends on the brand of paint used. the coverage amount is listed on the label of the paint can. use this to adjust the values in the paint coverage section below. paint usually is applied at 350 to 400 square feet per gallon (primer at 200 to 300 square feet per gallon).

step-by-step explanation: