{"id":191212,"date":"2025-02-14T09:50:29","date_gmt":"2025-02-14T09:50:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191212"},"modified":"2025-02-14T09:50:32","modified_gmt":"2025-02-14T09:50:32","slug":"use-the-ti-84-plus-calculator-to-find-the-z-scores-that-bound-the-middle-82-of-the-area-under-the-standard-normal-curve","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/14\/use-the-ti-84-plus-calculator-to-find-the-z-scores-that-bound-the-middle-82-of-the-area-under-the-standard-normal-curve\/","title":{"rendered":"Use the TI-84 Plus calculator to find the z-scores that bound the middle 82% of the area under the standard normal curve"},"content":{"rendered":"\n

Use the TI-84 Plus calculator to find the z-scores that bound the middle 82% of the area under the standard normal curve. Enter the answers in ascending and round to two decimal places. The z-scores for the given area are and X<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To find the z-scores that bound the middle 82% of the area under the standard normal curve using a TI-84 Plus calculator, you would typically follow these steps:<\/p>\n\n\n\n

Step-by-Step Guide:<\/h3>\n\n\n\n
    \n
  1. Understand the Problem<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • The problem asks for the z-scores that correspond to the middle 82% of the area under the standard normal curve.<\/li>\n\n\n\n
    • The standard normal curve has a mean of 0 and a standard deviation of 1.<\/li>\n\n\n\n
    • The area under the curve is symmetric, so the total area to the left and right of the center must add up to 82%. This means that 9% of the area is in each tail (since (100\\% – 82\\% = 18\\%), and (18\\%) divided by 2 gives 9% for each tail).<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Use the Inverse Normal Function<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • The area in each tail is 9%, so the cumulative area from the left to the z-score in question is 91% (since 100% – 9% = 91%).<\/li>\n\n\n\n
        • We will use the inverse normal<\/strong> function to find the z-scores for this cumulative area.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Using the TI-84 Plus<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • Press the 2nd<\/strong> button and then the VARS<\/strong> button to bring up the distribution menu.<\/li>\n\n\n\n
            • Select invNorm( )<\/strong>, which is the inverse normal function.<\/li>\n\n\n\n
            • For the z-score on the left side, enter the cumulative area of 0.09 (9% in the left tail).<\/li>\n\n\n\n
            • For the z-score on the right side, enter the cumulative area of 0.91 (91% cumulative area from the left tail). The inputs will look like this:<\/li>\n\n\n\n
            • invNorm(0.09, 0, 1)<\/strong> will give you the z-score for the left tail.<\/li>\n\n\n\n
            • invNorm(0.91, 0, 1)<\/strong> will give you the z-score for the right tail.<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Interpret the Results<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • The TI-84 Plus will output two z-scores. These z-scores will be the ones that bound the middle 82% of the standard normal curve.<\/li>\n<\/ul>\n\n\n\n

                  Solution:<\/h3>\n\n\n\n

                  After using the TI-84 Plus calculator, the z-scores for the 82% middle area are:<\/p>\n\n\n\n

                    \n
                  • Left z-score<\/strong>: approximately -1.41<\/strong><\/li>\n\n\n\n
                  • Right z-score<\/strong>: approximately 1.41<\/strong><\/li>\n<\/ul>\n\n\n\n

                    Explanation:<\/h3>\n\n\n\n
                      \n
                    • The middle 82% of the standard normal curve is symmetrically distributed around the mean of 0. Therefore, the two z-scores are equally distant from the mean, with one negative and one positive.<\/li>\n\n\n\n
                    • The 82% area corresponds to 9% in each tail of the curve, which means the z-scores that mark the boundaries of the central 82% area are -1.41<\/strong> (left) and 1.41<\/strong> (right).<\/li>\n\n\n\n
                    • These values are based on the inverse normal function, which gives us the z-scores for a specified cumulative probability under the normal curve.<\/li>\n<\/ul>\n\n\n\n

                      By using these z-scores, you can now understand the bounds for the middle 82% of the data in a normal distribution.<\/p>\n","protected":false},"excerpt":{"rendered":"

                      Use the TI-84 Plus calculator to find the z-scores that bound the middle 82% of the area under the standard normal curve. Enter the answers in ascending and round to two decimal places. The z-scores for the given area are and X The Correct Answer and Explanation is : To find the z-scores that bound […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-191212","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191212","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=191212"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191212\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191212"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191212"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}