Table continues in second photo! Thank you! ## Standard Normal Distribution Table (Z-Table)This table presents the cumulative probability associated with a particular Z-score in the context of a standard normal distribution. The Z-score represents the number of standard deviations a data point is from the mean of a distribution.To use the Z-table, follow these steps:1. **Identify the Z-score**: Determine the Z-score for your data point.2. **Locate the First Two Digits**: Find the first two digits of the Z-score in the leftmost column of the table (up to one decimal place).3. **Locate the Second Decimal Place**: Find the second decimal place of the Z-score in the topmost row of the table.4. **Find the Probability**: The intersection of the row and column provides the cumulative probability.### Z-Table Reference| z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 ||------|-------|-------|-------|-------|-------|-------|-------|-------|-------|-------|| 1.1 | .3643 | .3686 | .3729 | .3770 | .3810 | .3849 | .3888 | .3925 | .3962 | .3997 || 1.2 | .3849 | .3869 | .3888 | .3907 | .3925 | .3944 | .3962 | .3980 | .3997 | .4015 || 1.3 | .4032 | .4049 | .4066 | .4082 | .4099 | .4115 | .4131 | .4147 | .4162 | .4177 || 1.4 | .4192 | .4207 | .4222 | .4236 | .4251 | .4265 | .4279 | .4292 | .4306 | .4319 || 1.5 | .4332 | .4345 | .4357 | .4370 | .4382 | .4394 | .4406 | .4418 | .4429 | .4441 || 1.6 | .4452 | .4463 | .4474 | .4484 | .4495 | .4505 **Personnel Testing and Percentile Analysis**Personnel tests are designed to test a job applicant's cognitive and/or physical abilities. A particular dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last year, the distribution of scores was approximately normal with a mean of 78 and a standard deviation of 7.7.### Question Analysis:**a. A particular employer requires job candidates to score at least 84 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 84?****b. The testing service reported to a particular employer that one of its job candidates' scores fell at the 90th percentile of the distribution (i.e., approximately 90% of the scores were lower than the candidate's, and only 10% were higher). What was the candidate's score?**### Solution Steps:To solve the problems, we need to use the standard normal distribution table:**Step 1: Convert the Score to a Z-Score**\[ Z = \frac{(X - \mu)}{\sigma} \]Where:- $ X $ is the raw score (84 in question a)- $ \mu $ is the mean (78)- $ \sigma $ is the standard deviation (7.7)For question (a):\[ Z = \frac{(84 - 78)}{7.7} = \frac{6}{7.7} \approx 0.7792 \]Using the Z-Table to find the probability corresponding to the Z-score of 0.7792:\[ P(Z > 0.78) = 1 - P(Z \leq 0.78) \]From the Z-Table, $ P(Z \leq 0.78) \approx 0.7823 $:\[ P(Z > 0.78) = 1 - 0.7823 = 0.2177 \]So, approximately 21.8% of the test scores exceeded 84.**Step 2: Find the Candidate's Score at the 90th Percentile**From the Z-Table, the Z-score corresponding to the 90th percentile is approximately $ Z = 1.28 $.Convert the Z-score to the raw score:\[ X = \mu + Z \cdot \sigma \]\[ X = 78 +

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Description: Table continues in second photo! Thank you! ## Standard Normal Distribution Table (Z-Table)This table presents the cumulative probability associated with a particular Z-score in the context of a standard normal distribution. The Z-score represents th

From the given information, Consider, X is the score that follows normal distribution with mean of…

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MATLAB: An Introduction with Applications

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Chapter1: Starting With Matlab

Section: Chapter Questions

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Permutations and Combinations

If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.

Counting Theory

The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.

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Table continues in second photo! Thank you!

## Standard Normal Distribution Table (Z-Table)

This table presents the cumulative probability associated with a particular Z-score in the context of a standard normal distribution. The Z-score represents the number of standard deviations a data point is from the mean of a distribution.

To use the Z-table, follow these steps:

1. **Identify the Z-score**: Determine the Z-score for your data point.
2. **Locate the First Two Digits**: Find the first two digits of the Z-score in the leftmost column of the table (up to one decimal place).
3. **Locate the Second Decimal Place**: Find the second decimal place of the Z-score in the topmost row of the table.
4. **Find the Probability**: The intersection of the row and column provides the cumulative probability.

### Z-Table Reference

| z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
|------|-------|-------|-------|-------|-------|-------|-------|-------|-------|-------|
| 1.1 | .3643 | .3686 | .3729 | .3770 | .3810 | .3849 | .3888 | .3925 | .3962 | .3997 |
| 1.2 | .3849 | .3869 | .3888 | .3907 | .3925 | .3944 | .3962 | .3980 | .3997 | .4015 |
| 1.3 | .4032 | .4049 | .4066 | .4082 | .4099 | .4115 | .4131 | .4147 | .4162 | .4177 |
| 1.4 | .4192 | .4207 | .4222 | .4236 | .4251 | .4265 | .4279 | .4292 | .4306 | .4319 |
| 1.5 | .4332 | .4345 | .4357 | .4370 | .4382 | .4394 | .4406 | .4418 | .4429 | .4441 |
| 1.6 | .4452 | .4463 | .4474 | .4484 | .4495 | .4505

**Personnel Testing and Percentile Analysis**

Personnel tests are designed to test a job applicant's cognitive and/or physical abilities. A particular dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last year, the distribution of scores was approximately normal with a mean of 78 and a standard deviation of 7.7.

### Question Analysis:

**a. A particular employer requires job candidates to score at least 84 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 84?**

**b. The testing service reported to a particular employer that one of its job candidates' scores fell at the 90th percentile of the distribution (i.e., approximately 90% of the scores were lower than the candidate's, and only 10% were higher). What was the candidate's score?**

### Solution Steps:

To solve the problems, we need to use the standard normal distribution table:

**Step 1: Convert the Score to a Z-Score**

\[ Z = \frac{(X - \mu)}{\sigma} \]

Where:
- $ X $ is the raw score (84 in question a)
- $ \mu $ is the mean (78)
- $ \sigma $ is the standard deviation (7.7)

For question (a):

\[ Z = \frac{(84 - 78)}{7.7} = \frac{6}{7.7} \approx 0.7792 \]

Using the Z-Table to find the probability corresponding to the Z-score of 0.7792:

\[ P(Z > 0.78) = 1 - P(Z \leq 0.78) \]

From the Z-Table, $ P(Z \leq 0.78) \approx 0.7823 $:

\[ P(Z > 0.78) = 1 - 0.7823 = 0.2177 \]

So, approximately 21.8% of the test scores exceeded 84.

**Step 2: Find the Candidate's Score at the 90th Percentile**

From the Z-Table, the Z-score corresponding to the 90th percentile is approximately $ Z = 1.28 $.

Convert the Z-score to the raw score:

\[ X = \mu + Z \cdot \sigma \]

\[ X = 78 +

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