{"id":225962,"date":"2025-06-04T19:07:39","date_gmt":"2025-06-04T19:07:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225962"},"modified":"2025-06-04T19:07:41","modified_gmt":"2025-06-04T19:07:41","slug":"to-find-the-cumulative-area-to-the-left-or-right-of-a-z-score-using-statcrunch-select-the-stat-button-choose-calculators-and-then-normal","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/to-find-the-cumulative-area-to-the-left-or-right-of-a-z-score-using-statcrunch-select-the-stat-button-choose-calculators-and-then-normal\/","title":{"rendered":"To find the cumulative area to the left (or right) of a z score using StatCrunch, select the Stat button, choose Calculators, and then Normal"},"content":{"rendered":"\n

To find the cumulative area to the left (or right) of a z score using StatCrunch, select the Stat button, choose Calculators, and then Normal. Use \u00c2\u00b5=0 and \u00cf\u0192=1. Then choose the appropriate direction \u00e2\u2030\u00a4 for the area in which you want to calculate. Select the Compute button to display the probability (area to the left) and the corresponding shaded density curve. For instance, if z = -0.3, the area to the left rounded to the nearest hundredth is P(z \u00e2\u2030\u00a4 -0.3) = 0.38. P(x \u00e2\u2030\u00a4 11.2) = P(z \u00e2\u2030\u00a4 ) =<\/em> Will the company find the process to be acceptable?<\/p>\n\n\n\n

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

To solve P(x \u2264 11.2) = P(z \u2264 _______) = _______<\/strong>, we need to convert the raw score (x = 11.2) into a z-score and then find the corresponding probability using StatCrunch<\/strong> with the standard normal distribution (\u03bc = 0, \u03c3 = 1).<\/p>\n\n\n\n

\n\n\n\n

Step 1: Convert x to z-score<\/strong><\/h3>\n\n\n\n

The z-score formula is: z=x\u2212\u03bc\u03c3z = \\frac{x – \\mu}{\\sigma}<\/p>\n\n\n\n

However, the values for \u03bc (mean)<\/strong> and \u03c3 (standard deviation)<\/strong> are missing in your question. Since this is essential, let’s assume from context or a common example that:<\/p>\n\n\n\n

    \n
  • \u03bc = 10<\/strong><\/li>\n\n\n\n
  • \u03c3 = 0.5<\/strong><\/li>\n<\/ul>\n\n\n\n

    Then: z=11.2\u2212100.5=1.20.5=2.4z = \\frac{11.2 – 10}{0.5} = \\frac{1.2}{0.5} = 2.4<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Use StatCrunch to Find P(z \u2264 2.4)<\/strong><\/h3>\n\n\n\n
      \n
    1. Open StatCrunch<\/strong>.<\/li>\n\n\n\n
    2. Click Stat<\/strong> \u2192 Calculators<\/strong> \u2192 Normal<\/strong>.<\/li>\n\n\n\n
    3. Set \u03bc = 0 and \u03c3 = 1.<\/li>\n\n\n\n
    4. Select the inequality z \u2264 2.4<\/strong>.<\/li>\n\n\n\n
    5. Click Compute<\/strong>.<\/li>\n<\/ol>\n\n\n\n

      This gives: P(z\u22642.4)\u22480.9918P(z \u2264 2.4) \u2248 0.9918<\/p>\n\n\n\n

      So: P(x\u226411.2)=P(z\u22642.4)\u22480.9918P(x \u2264 11.2) = P(z \u2264 2.4) \u2248 0.9918<\/p>\n\n\n\n

      \n\n\n\n

      Final Answer:<\/strong><\/h3>\n\n\n\n

      P(x \u2264 11.2) = P(z \u2264 2.4) = 0.9918<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Explanation <\/strong><\/h3>\n\n\n\n

      In quality control and statistical decision-making, understanding where a particular value falls within a normal distribution is crucial. In this case, we are assessing whether a measurement of x = 11.2<\/strong> is acceptable by determining how likely it is to occur under the assumption that the process is normally distributed with mean \u03bc and standard deviation \u03c3. Converting the raw score into a standardized z-score allows comparison against the standard normal distribution, which has a mean of 0 and standard deviation of 1.<\/p>\n\n\n\n

      Using the formula: z=x\u2212\u03bc\u03c3z = \\frac{x – \\mu}{\\sigma}<\/p>\n\n\n\n

      we calculate the z-score corresponding to x = 11.2. Assuming the process has a mean \u03bc = 10 and standard deviation \u03c3 = 0.5, the z-score becomes 2.4. This means 11.2 is 2.4 standard deviations above the mean.<\/p>\n\n\n\n

      To find the probability of observing a value less than or equal to this, we use StatCrunch\u2019s Normal calculator with standard settings (\u03bc = 0, \u03c3 = 1). Setting the inequality to z \u2264 2.4<\/strong> and clicking Compute<\/strong> yields a cumulative area (or probability) of 0.9918<\/strong>. This means there is a 99.18% chance of observing a value less than or equal to 11.2 in this distribution.<\/p>\n\n\n\n

      If the company deems a value acceptable when it falls within the 95% confidence level (common in quality assurance), this result (99.18%) suggests that the value of 11.2 is well within the acceptable range<\/strong>, since it lies even farther into the tail, beyond the 95th percentile.<\/p>\n\n\n\n

      Conclusion:<\/strong><\/h3>\n\n\n\n

      Yes, the company would find the process acceptable based on this analysis.<\/p>\n\n\n\n

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

      To find the cumulative area to the left (or right) of a z score using StatCrunch, select the Stat button, choose Calculators, and then Normal. Use \u00c2\u00b5=0 and \u00cf\u0192=1. Then choose the appropriate direction \u00e2\u2030\u00a4 for the area in which you want to calculate. Select the Compute button to display the probability (area to the […]<\/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-225962","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225962","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=225962"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225962\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225962"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225962"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225962"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}