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

Answer:

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

Step-by-step 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 $\frac{1}{k^2}$ of teh data set lie outside the k standard deviations away from the mean i.e. at-least $(1-\frac{1}{k^2})$ of the distribution value fall within 'k' standard deviations.

So, $(1-\frac{1}{k^2})=0.9375$

$\frac{1}{k^2}=0.0625$

$k^2=\frac{1}{0.0625}$

$k^2=\frac{10000}{625}$

$k^2=16$

$k=4$

Length must be given by,

$L=(\mu-k\times \sigma,\mu+k\times \sigma)$

$L=(120-4\times 10,120+4\times 10)$

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

$L=(80,160)$

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