Answered: The patient recovery time from a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The patient recovery time from a surgical procedure is normally distributed with a mean of 6.3 days and a standard deviation of 5.1 days

1) What is the probability of spending more then two days in recovery?

2) What is the 80th precentile for recovery items? Interpret in a sentence.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The Z-score of a random variable X is defined as follows:

Z=(X–µ)/σ

Here, µ and σ are the mean and standard deviation of X, respectively.

Consider a random variable X that defines the recovery time.

According to the given information X follows normal distribution with mean 6.3 days and standard deviation 5.1 days.

The probability of spending more than two days in recovery is,

$\begin{array}{rcl} P (X & > & 2) = P (\frac{(X - µ)}{σ} > \frac{(2 - 6.3)}{5.1}) \\ = & 1 - P (Z < - 0.84) (U \sin g S \tan d a r d n o r m a l t a b l e : P (Z < - 0.84) = 0.2005) \\ = & 1 - 0.2005 \\ = & 0.7995 \end{array}$

Therefore, the probability of spending more than two days in recovery is 0.7995.

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