**Understanding Pregnancy Duration Using Z Scores**The lengths of pregnancies are typically normally distributed with a mean of 269 days and a standard deviation of 15 days. This educational example will walk you through finding the probability that a pregnancy will last 309 days or longer, and determining the length that separates premature babies from those who are not, using z-scores and the standard normal distribution table.**a. Finding the Probability of a Pregnancy Lasting 309 Days or Longer**To find the probability that a pregnancy will last 309 days or longer, use the z-score formula:\[ z = \frac{X - \mu}{\sigma} \]Where:- \( X \) is the value in question (309 days).- \( \mu \) is the mean (269 days).- \( \sigma \) is the standard deviation (15 days).Calculate the z-score:\[ z = \frac{309 - 269}{15} = \frac{40}{15} = 2.67 \]Using the z-score table, find the cumulative probability for \( z = 2.67 \). This corresponds to approximately 0.9963. To find the probability of the pregnancy lasting 309 days or longer, subtract this value from 1:\[ P(Z \geq 2.67) = 1 - P(Z \leq 2.67) = 1 - 0.9963 = 0.0037 \]Thus, the probability is approximately **0.0037** (rounded to four decimal places).**b. Determining the Length that Separates Premature Babies**Babies born in the lowest 4% are considered premature. Find the corresponding z-score for the 4th percentile (0.04) using the z-table. The z-score for 0.04 is approximately -1.75.Use the z-score formula to find the corresponding value (\( X \)):\[ -1.75 = \frac{X - 269}{15} \]Solve for \( X \):\[ X - 269 = -1.75 \times 15 \]\[ X - 269 = -26.25 \]\[ X = 269 - 26.25 \]\[ X = 242.75 \]Thus, babies born on or before **243 days** (rounded to the nearest integer) are considered premature.**Standard Normal Distribution Table Explanation # Understanding Positive z Scores## Standard Normal (z) Distribution: Cumulative Area from the LEFT### Overview The table below provides the cumulative area from the left for positive **z** scores in a standard normal distribution. This information is vital for those studying statistics, probability, and related fields, as it helps in determining the probability associated with a specific z value.### Table Explanation- **z**: Represents the z-score, which is the number of standard deviations a data point is from the mean.- The cumulative area represents the probability that a value is less than or equal to a particular z-score.### Detailed BreakdownThe table is divided into rows, each corresponding to a different z-score ranging from 0.0 to 2.5. #### Columns Explanation- **First Column (z)**: Lists the z-score.- **Other Columns (0.00 to 0.09)**: Represent the cumulative area values for the specific hundredths place. - For instance, for **z = 1.2**, the columns under **.00 to .09** provide cumulative areas for z = 1.20 to z = 1.29.#### Example UsageTo find the cumulative area from the left for a z-score of 1.3:1. Locate the row for **z = 1.3**.2. Find the column for **.00** (under 1.3 row), which is .9032.3. To find the area for **z = 1.34**, locate the row for 1.3 and then move to the column for .04, which corresponds to the area .9115.### Graph ExplanationThe diagram at the top displays a bell curve representing the standard normal distribution, with the mean at 0 and shaded area to the left of the marked z-score (**z**). This visually aids in understanding the area under the curve that corresponds to cumulative probabilities.### Full Table| z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 ||:---:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----

We have to find given probability.

The lengths of pregnancies are normally distributed with a mean of 269 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 premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. Standard Normal Table (Page 1) 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 NEGATIVE z Scores the nearest integer as needed.) Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 01 02 .03 .04 .05 .06 07 08 .09 -3.50

The lengths of pregnancies are normally distributed with a mean of 269 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 premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. Standard Normal Table (Page 1) 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 NEGATIVE z Scores the nearest integer as needed.) Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 01 02 .03 .04 .05 .06 07 08 .09 -3.50

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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