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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:

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Answer:

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.

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