{"id":234065,"date":"2025-06-13T17:34:53","date_gmt":"2025-06-13T17:34:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234065"},"modified":"2025-06-13T17:34:55","modified_gmt":"2025-06-13T17:34:55","slug":"z-is-a-standard-normal-variable","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/z-is-a-standard-normal-variable\/","title":{"rendered":"Z is a standard normal variable."},"content":{"rendered":"\n

Z is a standard normal variable. Determine the probability: The probability that z is between 1.18 and 1.97 P(1.18 < z < 1.97) = Round to four decimal places.<\/p>\n\n\n\n

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

Correct Answer:<\/strong>
P(1.18 < z < 1.97) = 0.1291<\/strong><\/p>\n\n\n\n

\n\n\n\n

Explanation:<\/strong><\/p>\n\n\n\n

In statistics, a standard normal variable (Z)<\/strong> refers to a random variable that follows the standard normal distribution \u2014 a normal distribution with a mean (\u03bc) of 0<\/strong> and a standard deviation (\u03c3) of 1<\/strong>. This distribution is symmetric and bell-shaped, and the probabilities associated with it can be found using standard normal distribution (Z) tables<\/strong> or statistical software.<\/p>\n\n\n\n

We are asked to find the probability that the Z value lies between 1.18 and 1.97<\/strong>, which is denoted as:P(1.18<Z<1.97)P(1.18 < Z < 1.97)P(1.18<Z<1.97)<\/p>\n\n\n\n

To solve this, we use the cumulative distribution function (CDF) of the standard normal distribution, denoted as \u03a6(z)<\/strong>. The cumulative probability up to a certain z-value gives the area under the curve to the left of that z-value.<\/p>\n\n\n\n

We calculate this probability by subtracting the cumulative probability up to 1.18 from the cumulative probability up to 1.97:P(1.18<Z<1.97)=P(Z<1.97)\u2212P(Z<1.18)P(1.18 < Z < 1.97) = P(Z < 1.97) – P(Z < 1.18)P(1.18<Z<1.97)=P(Z<1.97)\u2212P(Z<1.18)<\/p>\n\n\n\n

Using a standard normal table or calculator:<\/p>\n\n\n\n

    \n
  • \u03a6(1.97) \u2248 0.9756<\/li>\n\n\n\n
  • \u03a6(1.18) \u2248 0.8465<\/li>\n<\/ul>\n\n\n\n

    P(1.18<Z<1.97)=0.9756\u22120.8465=0.1291P(1.18 < Z < 1.97) = 0.9756 – 0.8465 = 0.1291P(1.18<Z<1.97)=0.9756\u22120.8465=0.1291<\/p>\n\n\n\n

    This means there is a 12.91%<\/strong> chance that a randomly selected value from the standard normal distribution will fall between 1.18 and 1.97.<\/p>\n\n\n\n

    The area under the normal curve between two z-scores represents the probability of the variable falling within that interval. Because the normal distribution is continuous and symmetric, this process applies universally for any z-values using the same subtraction method.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Z is a standard normal variable. Determine the probability: The probability that z is between 1.18 and 1.97 P(1.18 < z < 1.97) = Round to four decimal places. The Correct Answer and Explanation is: Correct Answer:P(1.18 < z < 1.97) = 0.1291 Explanation: In statistics, a standard normal variable (Z) refers to a random […]<\/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-234065","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234065","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=234065"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234065\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234065"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234065"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234065"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}