---**Title: Using the TI-84 Plus Calculator to Find Z-Scores****Objective:**Learn how to use the TI-84 Plus calculator to find the z-scores that bound the middle 96% of the area under the standard normal curve.**Instructions:**Use the TI-84 Plus calculator to find the z-scores that bound the middle 96% of the area under the standard normal curve. Once you find the z-scores, enter them in ascending order and round to two decimal places.**Question:**The z-scores for the given area are: [________] and [________]**Button and Functionality Explanation:**- **Input Boxes:** - Students are required to fill in two input boxes with their calculated z-scores. - The z-scores should be entered in ascending order. - Make sure to round the z-scores to two decimal places.- **Control Buttons:** - **"Continue" Button:** Click this button to proceed after entering the z-scores. - **"Submit Assignment" Button:** Once satisfied with the answers, click this button to submit the assignment.- **Icons on the Right Side:** - Represent options to view additional information or to reset the form if needed.**Footer:**A small footer with copyright information and associated links for terms of use and privacy policy.**Note:**This activity is set within an interactive educational software by McGraw Hill LLC.---**Graph/Diagram Explanation:**There are no graphs or diagrams displayed in this image.---

The middle 96% of the areaa under the standard nomal curve is 96%.

Answered: Use the TI-84 Plus calculator to find…

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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Suppose the lengths of human pregnancies are normally distributed with =266 days and o-16 days. Complete parts (a) and (b) below (a) The figure to the right represents the normal curve with u=266 days and e-16 days. The area to the right of X-300 is 0.0168. Provide two interpretations of this area Provide one interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice (Type integers or decimals) OA. The proportion of human pregnancies that last more than OB. The proportion of human pregnancies that last less than days is days is Select the correct choice below and fill in the answer boxes to complete your choice days is Provide a second interpretation of the area using the given values (Type integers or decimals) OA. The probability that a randomly selected human pregnancy lasts less than OB. The probability that a randomly selected human pregnancy lasts more than A 4 days is
Use the standard normal table to find the z-score that corresponds to the cumulative area 0.9616. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (Type an integer or decimal rounded to three decimal places as needed.) - X - X Standard normal table (page 1) Standard normal table (page 2) - 3.3 .0004 0005 .0003 .0004 .0004 .0004 .0004 .0004 .0005 .0005 .0005 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 - 3.2 .0005 .0005 .0006 .0006 .0006 .0006 .0006 .0007 .0007 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 3.1 .0007 .0007 .0008 .0008 .0008 .0008 .0009 .0009 .0009 .0010 0.4 .6554 .6591 .6628 .6700 .6736 .6772 .6808 .6844 .6879 .6664 - 3.0 – 2.9 .0010 .0010 .0011 .0011 .0011 .0012 .0012 .0013 .0013 .0013 .6915 .7019 .7088 .7422…
Use the TI-B4 calculator to find the z-score for which the area to its left is 0.12. Round the answer to two decimal places, The z-score for the given area is
Use the standard normal table to find the z-score that corresponds to the cumulative area 0.0129. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (Type an integer or decipaall eaundad to thcee deeisel elese apeaded - X Standard normal table (page 2) Standard normal table (page 1) Standard Normal Distribution Standard Normal Distribution Critical Values Critical Values Level of Confidence c Level of Confidence c 0.80 1.28 0.80 1.28 0.90 1.645 Area 0.90 1.645 0.95 1.96 0.95 1.96 0.99 2.575 0.99 2.575 .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .09 .08 .07 .06 05 .04 .03 .02 .01 .00 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 - 3.4 - 3.3 - 3.2 - 3.1 - 3.0 - 2.9 .0002 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 0.1 .5398 .5438 .5478 .5517…
can you please find the area of the left of 2.3451 for z score....
Use the standard normal table to find the area to the left of z=−0.34. The area to the left of z=−0.34 is
Find the z-score that best satisfies the condition. 10% of the total area is to the left of z. 2=0 (Round to the nearest hundredth.)
Use the standard normal table to find the z-score that corresponds to the cumulative area 0.9945 If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (Type an integer or decimal rounded to three decimal places as needed.)
Using the standard normal table in the Toolbox, determine a Z-value. Each response must include three numerical values and be rounded to the hundredths place. Question 1 The cumulative area to Z is 0.80. Question 2 The cumulative area to Z is 0.1762. Question 3 The area between Z and negative infinity is 0.6664. Question 4 The area between Z and positive infinity is 0.119.
The mean height of women in a country (ages 20 - 29) is 64.3 inches. A random sample of 70 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 inches? Assume o = 2.86. The probability that the mean height for the sample is greater than 65 inches is (Round to four decimal places as needed.)
Find the areas for the following z values. Use standard normal table. Provide your answers to four decimal places (4 dp.) (a) Area to the left of z= -0.25 - (b) Area to the right of z =1.23 - (c) Area to the right of z=-2.97 - (d) Area to the left of z=0.00 - (e) Area to the left of z=2.33 - ii) Using your calculator in STAT MODE, find the mean and standard deviation of the following sample for the variable age (Round to 3 dp). 18 18 20 20 23 25 29 35 Mean : Standard Deviation :

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Use the TI-84 Plus calculator to find the z-scores that bound the middle 96% of the area under the standard normal curve. Enter the answers in ascending order and round to two decimal places. The z-scores for the given area are and X S