p(z>-1.56) Find for the standard normal distribution. Table E The Standard Normal Distribution Cumulative Standard Normal Distribution .00 .01 .02 .03 .04 .05 .06 .07 .08 09 <-3.4 .0003 0003 0003 .0003 0003 0003 0003 0003 .0003 0002 <-3.3 0005 .0005 .0005 .0004 .0004 0004 .0004 .0004 .0004 .0003 <-3.2 0007 0007 0006 .0006 0006 0006 0006 0005 0005 .0005 <-3.1 J0010 0009 0009 .0009 .0008 .0008 .0008 .0008 0007 .0007 <-3.0 .0013 0013 .0013 .0012 0012 0011 0011 .0011 .0010 .0010 -2.9 0019 .0018 .0018 .0017 0016 .0016 .0015 0015 .0014 .0014 -2.8 0026 0025 .0024 .0023 .0023 0022 0021 0021 0020 .0019 -2.7 0035 0034 0033 .0032 0031 0030 .0029 .0028 .0027 .0026 -2.6 0047 .0045 0044 0043 0041 0040 0039 0038 .0037 0036 -2.5 0062 0060 0059 0057 .0055 0054 0052 0051 .0049 0048 -2.4 .0082 0080 .0078 .0075 .0073 0071 .0069 0068 .0066 0064 -2.3 0107 0104 0102 0099 0096 0094 .0091 0089 .0087 0084 -2.2 0139 0136 0132 0129 0125 0122 0119 .0116 0113 O110 -2.1 0179 0174 0170 0166 0162 0158 0154 0150 0146 0143 -2.0 .0228 0222 0217 0212 0207 0202 0197 0192 .0188 0183 -1.9 0287 0281 0274 0268 0262 0256 .0250 0244 .0239 .0233 -1.8 0359 0351 0344 0336 0329 0322 .0314 .0307 0301 0294 -1.7 .0446 0436 0427 0418 0409 0401 0392 .0384 0375 .0367 -1.6 0548 0537 0526 0516 0505 0495 .0485 .0475 0465 0455 -1.5 0668 0655 0643 0630 0618 0606 0594 0582 .0571 .0559 0.0594 O 0.9406 O 0.0156 O None of these
p(z>-1.56) Find for the standard normal distribution. Table E The Standard Normal Distribution Cumulative Standard Normal Distribution .00 .01 .02 .03 .04 .05 .06 .07 .08 09 <-3.4 .0003 0003 0003 .0003 0003 0003 0003 0003 .0003 0002 <-3.3 0005 .0005 .0005 .0004 .0004 0004 .0004 .0004 .0004 .0003 <-3.2 0007 0007 0006 .0006 0006 0006 0006 0005 0005 .0005 <-3.1 J0010 0009 0009 .0009 .0008 .0008 .0008 .0008 0007 .0007 <-3.0 .0013 0013 .0013 .0012 0012 0011 0011 .0011 .0010 .0010 -2.9 0019 .0018 .0018 .0017 0016 .0016 .0015 0015 .0014 .0014 -2.8 0026 0025 .0024 .0023 .0023 0022 0021 0021 0020 .0019 -2.7 0035 0034 0033 .0032 0031 0030 .0029 .0028 .0027 .0026 -2.6 0047 .0045 0044 0043 0041 0040 0039 0038 .0037 0036 -2.5 0062 0060 0059 0057 .0055 0054 0052 0051 .0049 0048 -2.4 .0082 0080 .0078 .0075 .0073 0071 .0069 0068 .0066 0064 -2.3 0107 0104 0102 0099 0096 0094 .0091 0089 .0087 0084 -2.2 0139 0136 0132 0129 0125 0122 0119 .0116 0113 O110 -2.1 0179 0174 0170 0166 0162 0158 0154 0150 0146 0143 -2.0 .0228 0222 0217 0212 0207 0202 0197 0192 .0188 0183 -1.9 0287 0281 0274 0268 0262 0256 .0250 0244 .0239 .0233 -1.8 0359 0351 0344 0336 0329 0322 .0314 .0307 0301 0294 -1.7 .0446 0436 0427 0418 0409 0401 0392 .0384 0375 .0367 -1.6 0548 0537 0526 0516 0505 0495 .0485 .0475 0465 0455 -1.5 0668 0655 0643 0630 0618 0606 0594 0582 .0571 .0559 0.0594 O 0.9406 O 0.0156 O None of these
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
Concept explainers
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
Question
100%
![### Finding \( P(z > -1.56) \) for the Standard Normal Distribution
Given the standard normal distribution, we aim to find \( P(z > -1.56) \). We will use the cumulative standard normal distribution table (Table E) provided in the image.
#### Table E: The Standard Normal Distribution
This table provides the cumulative probabilities for the standard normal distribution. The table displays the area to the left of a specified z-value.
##### Cumulative Standard Normal Distribution
The rows in the table represent the z-values up to the first decimal place, while the columns represent the hundredth decimal place of the z-values. For example:
- For \( z = -1.5 \), look at the row labeled \(-1.5\).
- For \( z = -1.56 \), look at the column labeled \(.06\) in the row for \(-1.5\).
Now follow these steps to find \( P(z > -1.56) \):
**Step 1: Identify \( P(z \leq -1.56) \)**
To find \( P(z \leq -1.56) \), locate the intersection of \( -1.5 \) row and the \( .06 \) column:
- The value at the intersection is \( 0.0594 \).
**Step 2: Calculate \( P(z > -1.56) \)**
Since the total area under the curve is 1, we use the complement rule:
\[ P(z > -1.56) = 1 - P(z \leq -1.56) \]
\[ P(z > -1.56) = 1 - 0.0594 \]
\[ P(z > -1.56) = 0.9406 \]
#### Explanation of the Diagram
The table shows \( P(z \leq z_0) \) for values of \( z_0 \) ranging from \(-3.4\) to \(3.4\). Each cell represents the cumulative probability up to that z-value. The rows and columns help pinpoint exact probabilities for various z-values to facilitate statistical analysis and problem-solving.
#### Conclusion
Therefore, \( P(z > -1.56) \) for the standard normal distribution is:
\[ 0.9406 \]
### Answer Options
- \( 0.0594 \)
- **\( 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfef55d1-e74e-4acc-a65e-b6e34bf1c85d%2F5f365316-734b-41f7-9048-6d1522148a0b%2Fikyebki_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding \( P(z > -1.56) \) for the Standard Normal Distribution
Given the standard normal distribution, we aim to find \( P(z > -1.56) \). We will use the cumulative standard normal distribution table (Table E) provided in the image.
#### Table E: The Standard Normal Distribution
This table provides the cumulative probabilities for the standard normal distribution. The table displays the area to the left of a specified z-value.
##### Cumulative Standard Normal Distribution
The rows in the table represent the z-values up to the first decimal place, while the columns represent the hundredth decimal place of the z-values. For example:
- For \( z = -1.5 \), look at the row labeled \(-1.5\).
- For \( z = -1.56 \), look at the column labeled \(.06\) in the row for \(-1.5\).
Now follow these steps to find \( P(z > -1.56) \):
**Step 1: Identify \( P(z \leq -1.56) \)**
To find \( P(z \leq -1.56) \), locate the intersection of \( -1.5 \) row and the \( .06 \) column:
- The value at the intersection is \( 0.0594 \).
**Step 2: Calculate \( P(z > -1.56) \)**
Since the total area under the curve is 1, we use the complement rule:
\[ P(z > -1.56) = 1 - P(z \leq -1.56) \]
\[ P(z > -1.56) = 1 - 0.0594 \]
\[ P(z > -1.56) = 0.9406 \]
#### Explanation of the Diagram
The table shows \( P(z \leq z_0) \) for values of \( z_0 \) ranging from \(-3.4\) to \(3.4\). Each cell represents the cumulative probability up to that z-value. The rows and columns help pinpoint exact probabilities for various z-values to facilitate statistical analysis and problem-solving.
#### Conclusion
Therefore, \( P(z > -1.56) \) for the standard normal distribution is:
\[ 0.9406 \]
### Answer Options
- \( 0.0594 \)
- **\( 0
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Recommended textbooks for you
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman