{"id":42954,"date":"2025-06-29T12:17:40","date_gmt":"2025-06-29T12:17:40","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42954"},"modified":"2025-06-29T12:17:42","modified_gmt":"2025-06-29T12:17:42","slug":"find-the-z-scores-for-which-28-of-the-distributions-area-lies-between-z-and-z","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-z-scores-for-which-28-of-the-distributions-area-lies-between-z-and-z\/","title":{"rendered":"Find the z-scores for which 28% of the distribution&#8217;s area lies between -z and z."},"content":{"rendered":"\n<p>Find the z-scores for which 28% of the distribution&#8217;s area lies between -z and z.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the z-scores for which 28% of the distribution\u2019s area lies between -z and z, we need to determine the z-scores corresponding to the cumulative probability that leaves 28% of the area in the middle of the normal distribution. This means that the total area in the tails of the distribution is 100% &#8211; 28% = 72%. Since the distribution is symmetric, the area in each tail is half of 72%, which is 36%.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step solution:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Area in the tails<\/strong>: The total area in the tails is 72%. Since the distribution is symmetric, the area in each tail is 36% (i.e., 72% \u00f7 2 = 36%).<\/li>\n\n\n\n<li><strong>Area to the left of z<\/strong>: To find the z-scores, we need to find the cumulative probability corresponding to the area to the left of z. The area between -z and z includes the central 28% of the distribution. Therefore, the cumulative area to the left of z will be: 50%+28%=78%50\\% + 28\\% = 78\\%50%+28%=78% This is because the total area to the left of -z is 50%, and adding the 28% area in the middle gives 78%.<\/li>\n\n\n\n<li><strong>Find the z-score<\/strong>: Now, we need to find the z-score that corresponds to the cumulative probability of 0.78 (or 78%) in the standard normal distribution. Using a standard normal distribution table or a z-score calculator, we find that a cumulative probability of 0.78 corresponds to a z-score of approximately <strong>0.77<\/strong>.<\/li>\n\n\n\n<li><strong>Conclusion<\/strong>: The z-scores for which 28% of the distribution&#8217;s area lies between -z and z are approximately <strong>-0.77 and 0.77<\/strong>.<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The standard normal distribution has a mean of 0 and a standard deviation of 1. Z-scores are a measure of how many standard deviations a data point is from the mean. For this problem, we used the cumulative area under the curve to find the z-scores that correspond to the middle 28% of the distribution. The value of 0.77 corresponds to the point where 78% of the distribution lies to the left of it. Thus, the z-scores are symmetric around 0, with the negative z-score being -0.77.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-401.jpeg\" alt=\"\" class=\"wp-image-42955\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-401.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-401-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-401-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-401-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the z-scores for which 28% of the distribution&#8217;s area lies between -z and z. The Correct Answer and Explanation is: To find the z-scores for which 28% of the distribution\u2019s area lies between -z and z, we need to determine the z-scores corresponding to the cumulative probability that leaves 28% of the area in [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42954","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42954","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42954"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42954\/revisions"}],"predecessor-version":[{"id":42956,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42954\/revisions\/42956"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42954"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42954"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42954"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}