Question

question 7
the time it takes a surgeon to complete a laparoscopic surgery to remove the gall bladder is normally distributed with a mean of 132.4 minutes and a standard deviation of 15.7 minutes. a patients risk of complications is increased the longer he is in surgery. an unusually risky surgery is one that is in the top 4% of all surgery lengths.
what is the minimum surgery length (in minutes) that would be considered unusually risky? note: answers are rounded to 1 decimal place.
a. 147.5 minutes
b. 247.4 minutes
c. 159.9 minutes
d. 104.9 minutes

Explanation:

Step1: Find the z - score

We know that an unusually risky surgery is in the top 4% of all surgery lengths. So the area to the left of the z - score is $1 - 0.04=0.96$. Looking up in the standard normal distribution table, the z - score corresponding to an area of 0.96 is approximately $z = 1.75$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value we want to find, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know that $\mu = 132.4$, $\sigma=15.7$, and $z = 1.75$. Rearranging the formula for $x$ gives $x=\mu + z\sigma$.

Step3: Calculate $x$

Substitute the values into the formula: $x=132.4+1.75\times15.7$. First, calculate $1.75\times15.7 = 1.75\times(15 + 0.7)=1.75\times15+1.75\times0.7=26.25 + 1.225 = 27.475$. Then $x=132.4+27.475 = 159.875\approx159.9$.

Answer:

C. 159.9 minutes