A hospital researcher has found that the mean length of time that

[roam]

A hospital researcher has found that the mean length of time that patients stay in the hospital is 4.9 days with a standard deviation of 3.5 days.

a) Using just the information on the mean and standard deviation, justify why the length of stay would NOT be normally distributed. Would the length of stay be positively skewed or negatively skewed?

So, what is the mode or median for this? Because I need to use the Pearson's formulas and we need to know either the mode/median.

Skewness = (mean - mode / standard deviation)
Skewness = 3(mean - median / standard deviation)

Skewness = $$\frac{\bar{x} - mode}{s}$$ or
$$\frac{3(\bar{x} - median)}{s}$$

 

[mathman]

Since the length of a hospital stay has to be non-negative, it can't be normally distributed. The upper end is stretched out, could be more than 9.8 days. As a result the mean would be greater than the median.

 

[roam]

Therefore the length of stay would be positively skewed?

 

[mathman]

Therefore the length of stay would be positively skewed?

Yes! Why did you ask??

 

[statdad]

Think about the basic properties of any normal distribution: symmetry about $$\mu$$, and the fact that the two sides extend approximately three standard deviations away from the mean. In this case, for this mean and standard deviation, the left end of the curve would extend to the left of zero, indicating a negative length of hospital stay which, as has been pointed out, can't happen.

From an intuitive point of view, while most people are not hospitalized any great length of time, a few stay for quite a while - times are right-skewed.

 

[BWV]

FWIW the poisson and exponential distributions are most commonly used for cueing problems

The other option would be lognormal where the log of the length of hospital stay would be normally distributed, this would produce a skewed distribution as the log cannot be negative

 

[statdad]

"FWIW the poisson and exponential distributions are most commonly used for cueing problems"

True, the Poisson distribution is often used to model the pattern of arrivals, and the exponential to model the inter-arrival times, but this does not seem to be one of those problems - the OP is not discussing the pattern of arrivals, or waiting times (unless there is more to post).

 

[CRGreathouse]

In this case, for this mean and standard deviation, the left end of the curve would extend to the left of zero, indicating a negative length of hospital stay which, as has been pointed out, can't happen.

Back in *my* day, we had to walk three miles to school in the freezing rain and stay in hospitals for a negative amount of time.

 

[statdad]

"Back in *my* day, we had to walk three miles to school in the freezing rain and stay in hospitals for a negative amount of time."

Be thankful you were able to stay. Where I grew up we had to walk through freezing rain and hail, UPHILL, to the hospital, start the furnace by rubbing icicles together, then walk home, UPHILL, through a snowstorm, to get home. If we didn't arrive before we left we were punished.

 

[HallsofIvy]

You had a home? Boy, you were really lucky!

 

[statdad]

"You had a home? Boy, you were really lucky!"

Well, it was really our neighbor's home, a box, and the walkway to their door was lined with broken glass. That wouldn't have been so bad, but we had to pay them to take our shoes and socks before they would grant us entrance.

 

[ssd]

IMO, you cannot assume normality because mean-k*sigma is negative for k>1.4. For X~N(mu,sigma), P( X<mu-1.4*sigma) is not negligible (0.081 approx). But here this probability is 0.