{"id":12639,"date":"2022-12-20T22:31:53","date_gmt":"2022-12-20T17:01:53","guid":{"rendered":"https:\/\/mathemerize.com\/?p=12639"},"modified":"2022-12-20T22:31:56","modified_gmt":"2022-12-20T17:01:56","slug":"the-lengths-of-40-leaves-of-an-plant-are-measured-correct-to-the-nearest-millimetre-and-the-data-obtained-is-represented-in-the-following-table-find-the-median-length-of-the-leaves","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-lengths-of-40-leaves-of-an-plant-are-measured-correct-to-the-nearest-millimetre-and-the-data-obtained-is-represented-in-the-following-table-find-the-median-length-of-the-leaves\/","title":{"rendered":"The lengths of 40 leaves of an plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves."},"content":{"rendered":"\n

Question :<\/h2>\n\n\n\n
Length (in mm)<\/td>118 – 126<\/td>127 – 135<\/td>136 – 144<\/td>145 – 153<\/td>154 – 162<\/td>163 – 171<\/td>172 – 180<\/td><\/tr>
Number of Leaves<\/td>3<\/td>5<\/td>9<\/td>12<\/td>5<\/td>4<\/td>2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution :<\/h2>\n\n\n\n

Here the frequency table is given in the inclusive form. So, we first convert it into exclusive form by subtracting and adding \\(h\\over 2\\) to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and the upper limit of the previous class.<\/p>\n\n\n\n

Converting the given table into exclusive form and preparing the cumulative frequency table, we get<\/p>\n\n\n\n

Length (in mm)<\/td>Number of leaves (frequency)<\/td>Cumulative frequency<\/td><\/tr>
117.5 – 126.5<\/td>3<\/td>3<\/td><\/tr>
126.5 – 135.5<\/td>5<\/td>8<\/td><\/tr>
135.5 – 144.5<\/td>9<\/td>17<\/td><\/tr>
144.5 – 153.5<\/td>12<\/td>29<\/td><\/tr>
153.5 – 162.5<\/td>5<\/td>34<\/td><\/tr>
162.5 – 171.5<\/td>4<\/td>38<\/td><\/tr>
171.5 – 180.5<\/td>2<\/td>40<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

We have : n = 40 So, \\(n\\over 2\\) = 20<\/p>\n\n\n\n

The cumulative frequency just greater than \\(n\\over 2\\) is 29 and the corresponding class is (144.5 – 153.5).<\/p>\n\n\n\n

So, it is the median class.<\/p>\n\n\n\n

Here, l = 144.5, h = 9, f = 12 and cf = 17<\/p>\n\n\n\n

Substituting these values in the formula<\/p>\n\n\n\n

Median = l + (\\({n\\over 2} – cf\\over f\\) \\(\\times\\) h,<\/p>\n\n\n\n

Median = 144.5 + \\(3\\over 12\\) \\(\\times\\) 9 = 144.5 + 2.25 = 146.75 mm<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Length (in mm) 118 – 126 127 – 135 136 – 144 145 – 153 154 – 162 163 – 171 172 – 180 Number of Leaves 3 5 9 12 5 4 2 Solution : Here the frequency table is given in the inclusive form. So, we first convert it into exclusive …<\/p>\n

The lengths of 40 leaves of an plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,58],"tags":[],"yoast_head":"\nThe lengths of 40 leaves of an plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. 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