POSITIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 5279 .5319 .5359 0.1 .5398 5438 5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 5832 5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 6844 .6879 0,5 .6915 6950 ,6985 7019 .7054 .7088 .7123 7157 .7190 .7224 0.6 .7257 7291 7324 .7357 7389 7422 .7454 .7486 .7517 .7549 0.7 7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 7823 .7852 0.8 .7881 7910 7939 7967 7995 .8023 .8051 .8078 .8106 .8133 0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 8485 .8508 .8531 .8554 .8577 ,8599 .8621 1.1 .8643 8665 8686 .8708 .8729 .8749 .8770 ,8790 .8810 .8830 1.2 8849 .8869 8888 .8907 .8925 8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 9474 9484 9495 .9505 ,9515 .9525 ,9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 9010 0025 9927 9071 0074
POSITIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 5279 .5319 .5359 0.1 .5398 5438 5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 5832 5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 6844 .6879 0,5 .6915 6950 ,6985 7019 .7054 .7088 .7123 7157 .7190 .7224 0.6 .7257 7291 7324 .7357 7389 7422 .7454 .7486 .7517 .7549 0.7 7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 7823 .7852 0.8 .7881 7910 7939 7967 7995 .8023 .8051 .8078 .8106 .8133 0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 8485 .8508 .8531 .8554 .8577 ,8599 .8621 1.1 .8643 8665 8686 .8708 .8729 .8749 .8770 ,8790 .8810 .8830 1.2 8849 .8869 8888 .8907 .8925 8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 9474 9484 9495 .9505 ,9515 .9525 ,9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 9010 0025 9927 9071 0074
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Transcribed Image Text:## Standard Normal (z) Distribution: Cumulative Area from the LEFT for Positive z Scores
### Explanation of the Graph
The image includes a normal distribution curve where the shaded area to the left of a vertical line at a value z represents the cumulative probability up to that z-score. This cumulative area can be used to find probabilities and percentile ranks associated with z-scores in a standardized normal distribution.
### Positive z Scores Table
This table provides cumulative probabilities for positive z-scores. The z-scores are listed in the leftmost column, and the corresponding probabilities are presented in subsequent columns. Each z-score value corresponds to the area under the normal distribution curve to the left of that z-score.
| z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
|------|--------|--------|--------|--------|--------|--------|--------|--------|--------|--------|
| 0.0 | .5000 | .5040 | .5080 | .5120 | .5160 | .5199 | .5239 | .5279 | .5319 | .5359 |
| 0.1 | .5398 | .5438 | .5478 | .5517 | .5557 | .5596 | .5636 | .5675 | .5714 | .5753 |
| 0.2 | .5793 | .5832 | .5871 | .5910 | .5948 | .5987 | .6026 | .6064 | .6103 | .6141 |
| 0.3 | .6179 | .6217 | .6255 | .6293 | .6331 | .6368 | .6406 | .6443 | .6480 | .6517 |
| 0.4 | .6554 | .6591 | .6628 | .6664 | .6700 | .6736 | .6772 | .6808 | .6844 | .6879 |
| 0.5 | .
![## Understanding Normal Distribution for IQ Scores
### Problem Statement
Assume that adults have IQ scores that are normally distributed with a mean (μ) of 105 and a standard deviation (σ) of 20. Find the probability that a randomly selected adult has an IQ between 89 and 121.
### Steps to Solve
1. **Convert the IQ scores to z-scores:**
- Calculating the z-score for IQ = 89:
\[
z = \frac{X - \mu}{\sigma} = \frac{89 - 105}{20} = -0.8
\]
- Calculating the z-score for IQ = 121:
\[
z = \frac{X - \mu}{\sigma} = \frac{121 - 105}{20} = 0.8
\]
2. **Using the Standard Normal Table:**
- The provided standard normal distribution table lists the cumulative area from the left for negative z-scores.
- Locate z = -0.8 in the table:
- The intersection of the row and column for -0.8 gives the cumulative probability for IQ < 89.
- The table provides area values for z from -3.4 to 0.0 with increments of 0.1.
### Detailed Description of the Graph/Diagram
The image includes a section of the "Standard Normal Table (Page 1)" focused on **Negative z Scores**. The table presents:
- **Rows labeled as \( z \)**: Indicate the z-score values.
- **Columns labeled from .01 to .09**: Represent the decimal parts of the z-score.
- **Cell values**: Each cell contains the cumulative probability from the left up to the z-score formed by combining the row and column values.
**Example:**
- For \( z = -1.5 \) and a column value of .07, find the intersection which gives a cumulative probability around 0.0668.
### Answer Box
Type your final answer in terms of probability rounded to four decimal places in the provided answer box.
### Hint Link
- **Click to view page 1 of the table.**
- **Click to view page 2 of the table.**
### Summary
By using the z-scores and the standard normal distribution table, you can find the probability that a randomly selected adult has an IQ between any](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e4f48b8-ad6a-4fd5-bfc6-01337edf5aa9%2F99dbdaf0-c3f7-4caa-bdb1-65f20ac92e29%2Fzkoszv_processed.png&w=3840&q=75)
Transcribed Image Text:## Understanding Normal Distribution for IQ Scores
### Problem Statement
Assume that adults have IQ scores that are normally distributed with a mean (μ) of 105 and a standard deviation (σ) of 20. Find the probability that a randomly selected adult has an IQ between 89 and 121.
### Steps to Solve
1. **Convert the IQ scores to z-scores:**
- Calculating the z-score for IQ = 89:
\[
z = \frac{X - \mu}{\sigma} = \frac{89 - 105}{20} = -0.8
\]
- Calculating the z-score for IQ = 121:
\[
z = \frac{X - \mu}{\sigma} = \frac{121 - 105}{20} = 0.8
\]
2. **Using the Standard Normal Table:**
- The provided standard normal distribution table lists the cumulative area from the left for negative z-scores.
- Locate z = -0.8 in the table:
- The intersection of the row and column for -0.8 gives the cumulative probability for IQ < 89.
- The table provides area values for z from -3.4 to 0.0 with increments of 0.1.
### Detailed Description of the Graph/Diagram
The image includes a section of the "Standard Normal Table (Page 1)" focused on **Negative z Scores**. The table presents:
- **Rows labeled as \( z \)**: Indicate the z-score values.
- **Columns labeled from .01 to .09**: Represent the decimal parts of the z-score.
- **Cell values**: Each cell contains the cumulative probability from the left up to the z-score formed by combining the row and column values.
**Example:**
- For \( z = -1.5 \) and a column value of .07, find the intersection which gives a cumulative probability around 0.0668.
### Answer Box
Type your final answer in terms of probability rounded to four decimal places in the provided answer box.
### Hint Link
- **Click to view page 1 of the table.**
- **Click to view page 2 of the table.**
### Summary
By using the z-scores and the standard normal distribution table, you can find the probability that a randomly selected adult has an IQ between any
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