Question

2. Hospital records show that a certain surgical procedure takes on the average of 120 minutes with a standard deviation of 10 minutes. Between how many minutes must be the lengths of at least 93.75% of these surgical procedures? Answer:

Answer 1

93.75% of these surgical procedures takes at-least 80 to 160 minutes.

Explanation:

Given : Hospital records show that a certain surgical procedure takes on the average of 120 minutes with a standard deviation of 10 minutes.

To find : Between how many minutes must be the lengths of at least 93.75% of these surgical procedures?

Solution :

The average mean is
`\mu=120` minutes.

The standard deviation is
`\sigma=10` minutes.

At least 93.75% of these surgical procedures,

Applying Chebyshev's theorem,

For any constant k>1, no more than
`(1)/(k^2)` of teh data set lie outside the k standard deviations away from the mean i.e. at-least
`(1-(1)/(k^2))` of the distribution value fall within 'k' standard deviations.

So,
`(1-(1)/(k^2))=0.9375`

`(1)/(k^2)=0.0625`

`k^2=(1)/(0.0625)`

`k^2=(10000)/(625)`

`k^2=16`

`k=4`

Length must be given by,

`L=(\mu-k* \sigma,\mu+k* \sigma)`

`L=(120-4* 10,120+4* 10)`

`L=(120-40,120+40)`

`L=(80,160)`

Therefore, 93.75% of these surgical procedures takes at-least 80 to 160 minutes.