{"id":12635,"date":"2022-12-20T19:43:56","date_gmt":"2022-12-20T14:13:56","guid":{"rendered":"https:\/\/mathemerize.com\/?p=12635"},"modified":"2022-12-20T19:45:21","modified_gmt":"2022-12-20T14:15:21","slug":"a-life-insurance-agent-found-the-following-data-for-distribution-of-ages-of-100-policy-holders-calculate-the-median-age-if-policies-are-given-only-to-persons-having-age-18-years-onwards-but-less-tha","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-life-insurance-agent-found-the-following-data-for-distribution-of-ages-of-100-policy-holders-calculate-the-median-age-if-policies-are-given-only-to-persons-having-age-18-years-onwards-but-less-tha\/","title":{"rendered":"A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years."},"content":{"rendered":"\n

Question :<\/h2>\n\n\n\n
Age (in years)<\/td>Number of Policy Holders<\/td><\/tr>
Below 20<\/td>2<\/td><\/tr>
Below 25<\/td>6<\/td><\/tr>
Below 30<\/td>24<\/td><\/tr>
Below 35<\/td>45<\/td><\/tr>
Below 40<\/td>78<\/td><\/tr>
Below 45<\/td>89<\/td><\/tr>
Below 50<\/td>92<\/td><\/tr>
Below 55<\/td>98<\/td><\/tr>
Below 60<\/td>100<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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

We are given the cumulative frequency distribution.<\/p>\n\n\n\n

So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median.<\/p>\n\n\n\n

Class Interval<\/td>Frequency<\/td>Cumulative Frequency<\/td><\/tr>
18 – 20<\/td>2<\/td>2<\/td><\/tr>
20 – 25<\/td>4<\/td>6<\/td><\/tr>
25 – 30<\/td>18<\/td>24<\/td><\/tr>
30 – 35<\/td>21<\/td>45<\/td><\/tr>
35 – 40<\/td>33<\/td>78<\/td><\/tr>
40 – 45<\/td>11<\/td>89<\/td><\/tr>
45 – 50<\/td>3<\/td>92<\/td><\/tr>
50 – 55<\/td>6<\/td>98<\/td><\/tr>
55 – 60<\/td>2<\/td>100<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Here, n = 100 So, \\(n\\over 2\\) = 50<\/p>\n\n\n\n

We see that the cumulative frequency just greater than \\(n\\over 2\\), i.e. 50 is 78 and the corresponding class is (35 – 40), So, it is the median class.<\/p>\n\n\n\n

\\(\\therefore\\) l = 35, cf = 45, f = 33 and h = 5.<\/p>\n\n\n\n

Now, let us substitute these values in the formula<\/p>\n\n\n\n

Median = l + (\\({n\\over 2} – cf\\over f\\)) \\(\\times\\) h = 35 + \\(5\\over 33\\) \\(\\times\\) 5<\/p>\n\n\n\n

= 35 + 0.76 = 35.76<\/p>\n\n\n\n

Hence, the median age is 35.76<\/strong> years.<\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Age (in years) Number of Policy Holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100 Solution : We are given the cumulative frequency distribution. So, we first construct a frequency table from the given …<\/p>\n

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.<\/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":"\nA life insurance agent found the following data for distribution of ages of 100 policy holders. 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