Question #4365

The average length of a hospital stay for all diagnoses is 4.8 days. If we assume that the lengths of hospital stays are normally distributed with a variance of 2.1, then 10% of hospital stays are longer than how many days? Thirty percent of stays are less than how many days?

Expert's answer

First let's calculate the standard deviation.

Variance = (standard deviation)^2

Standard deviation = sqrt(2.1)

Standard Deviation = 1.45

Now we must use z-scores. To find the top 10% of hospitals, we must find the z-score that corresponds to 90%. Looking at a z-score table, we look for 0.9000 and can find this z-score to be about 1.28. Then with the formula for z-scores, we can find the corresponding value.

z = (x-μ)/σ

1.28 = (x-4.8)/1.45

x = 6.656

Thus 10% of hospital stays are longer than 6.656 days.

To find the lowest 30% we can look for 0.3000 and can find this z-score to be about -0.525. Then again with the z-score formula we have:

-0.525 = (x-4.8)/1.45

x = 4.03875

Thus 30% of stays are less than 4.03875 days.

Variance = (standard deviation)^2

Standard deviation = sqrt(2.1)

Standard Deviation = 1.45

Now we must use z-scores. To find the top 10% of hospitals, we must find the z-score that corresponds to 90%. Looking at a z-score table, we look for 0.9000 and can find this z-score to be about 1.28. Then with the formula for z-scores, we can find the corresponding value.

z = (x-μ)/σ

1.28 = (x-4.8)/1.45

x = 6.656

Thus 10% of hospital stays are longer than 6.656 days.

To find the lowest 30% we can look for 0.3000 and can find this z-score to be about -0.525. Then again with the z-score formula we have:

-0.525 = (x-4.8)/1.45

x = 4.03875

Thus 30% of stays are less than 4.03875 days.

## Comments

Assignment Expert01.07.15, 18:36Dear Anon.

Both variance and standard deviation are applied in statistics . Besides, you can see all calculations involve a standard deviation, not variance.

Anon.01.07.15, 08:57That damn variance got me. Forgot to turn it into a standard deviation!

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