Many sharks enter a state of tonic immobility when inverted. Suppose that in a particular species of sharksthe time a shark remains in a state of tonic immobility when inverted is normally distributed with mean 11.2minutes and standard deviation 1.1 minutes.<\/p>\n\n\n\n
a. If a biologist induces a state of tonic immobility in such a shark in order to study it, find theprobability that the shark will remain in this state for between 10 and 13 minutes.<\/p>\n\n\n\n
b. When a biologist wishes to estimate the mean time that such sharks stay immobile by inducing tonic immobility in each of a sample of 12 sharks, find the probability that mean time ofimmobility in the sample will be between 10 and 13 minutes.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n In this part, we are given that the time a shark remains in a state of tonic immobility is normally distributed with a mean of 11.2 minutes and a standard deviation of 1.1 minutes. We are asked to find the probability that the shark remains in this state for between 10 and 13 minutes.<\/p>\n\n\n\n To solve this, we can use the standard normal distribution by first transforming the raw times (10 and 13 minutes) into Z-scores using the formula:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n For ( X = 10 ) minutes:<\/p>\n\n\n\n [ For ( X = 13 ) minutes:<\/p>\n\n\n\n [ From the standard normal distribution table:<\/p>\n\n\n\n The probability that the shark remains in tonic immobility for between 10 and 13 minutes is the difference between the cumulative probabilities:<\/p>\n\n\n\n [ So, the probability is approximately 0.8116<\/strong> or 81.16%<\/strong>.<\/p>\n\n\n\n In this part, we are considering the mean time of immobility for a sample of 12 sharks. The sampling distribution of the sample mean will follow a normal distribution with:<\/p>\n\n\n\n [ For ( X = 10 ) minutes:<\/p>\n\n\n\n [ For ( X = 13 ) minutes:<\/p>\n\n\n\n [ From the standard normal distribution table:<\/p>\n\n\n\n [ So, the probability is approximately 0.9999<\/strong> or 99.99%<\/strong>.<\/p>\n\n\n\n The key difference between parts (a) and (b) lies in the sampling process. In part (a), we are looking at the individual time of immobility for a single shark, which follows a normal distribution with a mean of 11.2 minutes and a standard deviation of 1.1 minutes. In part (b), we are considering the sample mean of 12 sharks, which has a smaller standard deviation (standard error) because we average the times. This results in a much higher probability that the sample mean will fall between 10 and 13 minutes, reflecting the fact that averages tend to be less variable than individual observations.<\/p>\n","protected":false},"excerpt":{"rendered":" Many sharks enter a state of tonic immobility when inverted. Suppose that in a particular species of sharksthe time a shark remains in a state of tonic immobility when inverted is normally distributed with mean 11.2minutes and standard deviation 1.1 minutes. a. If a biologist induces a state of tonic immobility in such a shark […]<\/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-200023","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/200023","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=200023"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/200023\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=200023"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=200023"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=200023"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part (a) \u2014 Probability that a shark remains in tonic immobility for between 10 and 13 minutes:<\/h3>\n\n\n\n
Z = \\frac{X – \\mu}{\\sigma}
]<\/p>\n\n\n\n\n
Step 1: Calculate the Z-scores<\/h4>\n\n\n\n
Z_1 = \\frac{10 – 11.2}{1.1} = \\frac{-1.2}{1.1} \\approx -1.09
]<\/p>\n\n\n\n
Z_2 = \\frac{13 – 11.2}{1.1} = \\frac{1.8}{1.1} \\approx 1.64
]<\/p>\n\n\n\nStep 2: Find the probabilities corresponding to these Z-scores using the standard normal distribution table.<\/h4>\n\n\n\n
\n
Step 3: Calculate the probability between these Z-scores:<\/h4>\n\n\n\n
P(10 \\leq X \\leq 13) = P(Z_2) – P(Z_1) = 0.9495 – 0.1379 = 0.8116
]<\/p>\n\n\n\nPart (b) \u2014 Probability that the mean time of immobility in a sample of 12 sharks is between 10 and 13 minutes:<\/h3>\n\n\n\n
\n
Step 1: Calculate the standard error of the mean:<\/h4>\n\n\n\n
\\sigma_{\\text{mean}} = \\frac{1.1}{\\sqrt{12}} \\approx \\frac{1.1}{3.464} \\approx 0.318
]<\/p>\n\n\n\nStep 2: Standardize the values (find the Z-scores):<\/h4>\n\n\n\n
Z_1 = \\frac{10 – 11.2}{0.318} = \\frac{-1.2}{0.318} \\approx -3.77
]<\/p>\n\n\n\n
Z_2 = \\frac{13 – 11.2}{0.318} = \\frac{1.8}{0.318} \\approx 5.66
]<\/p>\n\n\n\nStep 3: Find the cumulative probabilities for these Z-scores:<\/h4>\n\n\n\n
\n
Step 4: Calculate the probability between these Z-scores:<\/h4>\n\n\n\n
P(10 \\leq \\bar{X} \\leq 13) = P(Z_2) – P(Z_1) = 1.0000 – 0.00008 = 0.99992
]<\/p>\n\n\n\nSummary of Results:<\/h3>\n\n\n\n
\n
Explanation:<\/h3>\n\n\n\n