{"id":20330,"date":"2024-04-15T05:51:26","date_gmt":"2024-04-15T05:51:26","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20330"},"modified":"2024-04-15T05:51:26","modified_gmt":"2024-04-15T05:51:26","slug":"the-diameters-of-10000-ball-bearings-were-measured-the-mean-diameter-and-standard-deviation-were-found-to-be-10-mm-and-0-05-mm-respectively-assuming-gaussian-distribution-of-measurements-it-can-be","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-diameters-of-10000-ball-bearings-were-measured-the-mean-diameter-and-standard-deviation-were-found-to-be-10-mm-and-0-05-mm-respectively-assuming-gaussian-distribution-of-measurements-it-can-be\/","title":{"rendered":"The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05 mm respectively. Assuming Gaussian distribution of measurements, it can be expected that the number of measurement more than 10.15 mm will be A. 230 B. 115 C. 15 D. 2"},"content":{"rendered":"<p>[amp_mcq option1=&#8221;230&#8243; option2=&#8221;115&#8243; option3=&#8221;15&#8243; option4=&#8221;2&#8243; correct=&#8221;option1&#8243;]<!--more--><\/p>\n<p>The correct answer is $\\boxed{\\text{B) 115}}$.<\/p>\n<p>The number of measurements more than 10.15 mm is equal to the area under the Gaussian curve to the right of 10.15 mm. We can use the z-score to calculate this area. The z-score is calculated by subtracting the mean from the value and then dividing by the standard deviation. In this case, the z-score is 0.3, which corresponds to an area of 0.6179 under the curve. Multiplying this area by the total number of measurements (10000) gives us 115.<\/p>\n<p>Option A is incorrect because it is the area under the curve to the right of 10.3, which is a larger value than 10.15 mm. Option C is incorrect because it is the area under the curve to the right of 10.1, which is a smaller value than 10.15 mm. Option D is incorrect because it is the area under the curve to the right of 10.05, which is a very small value.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[amp_mcq option1=&#8221;230&#8243; option2=&#8221;115&#8243; option3=&#8221;15&#8243; option4=&#8221;2&#8243; correct=&#8221;option1&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[691],"tags":[],"class_list":["post-20330","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05 mm respectively. Assuming Gaussian distribution of measurements, it can be expected that the number of measurement more than 10.15 mm will be A. 230 B. 115 C. 15 D. 2<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-diameters-of-10000-ball-bearings-were-measured-the-mean-diameter-and-standard-deviation-were-found-to-be-10-mm-and-0-05-mm-respectively-assuming-gaussian-distribution-of-measurements-it-can-be\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05 mm respectively. 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