Answered: a. The probability that a pregnancy…

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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Question

The lengths of pregnancies are normally distributed with a mean of 268
days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest 2%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

a. The probability that a pregnancy will last 309 days or longer is

(Round to four decimal places as needed.)

b. Babies who are born on or before____days are considered premature.

(Round to the nearest integer as needed.)

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

a. The probability that a pregnancy will last 309 days or longer is

(Round to four decimal places as needed.)

The following information has been provided

$\mu = 268, \text{ } \sigma = 15$

We need to compute $\Pr(X \ge 309)$

The corresponding z-value needed to be computed is:

$Z = \frac{X - \mu}{\sigma} = \frac{ 309 - 268}{ 15} = 2.7333$

Therefore, we get that

$\Pr(X \geq 309) = \Pr\left(Z \ge \frac{ 309 - 268}{ 15}\right) = \Pr(Z \ge 2.7333)$

$= 1 - 0.9969 = 0.0031$

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