Question

the class marks of a distribution are 37 42 and 47 the class limits corresponding to class mark 42 are

Answer 1

`39.5,44.5`

Explanation:

`Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ first\ interval\ be\ u_1,v_1.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ second\ interval\ be\ u_2,v_2.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ third\ interval\ be\ u_3,v_3.\\Hence,\\As\ the\ class\ marks\ are\ uniform\ with\ an\ a\ difference\ of\ 5, the\\ observation\ is\ continuous\ and\ the\ class\ size\ is\ 5\ too.\\Hence,\\v_1=u_2, v_2=u_3\\`
`Now,\\Lets\ consider\ the\ first\ two\ classes.\\37=(u_1+v_1)/(2) \\42=(u_2+v_2)/(2)\\By\ adding\ the\ equations:\\79= (u_1+v_1)/(2)+(u_2+v_2)/(2)\\79=(u_1+v_1+u_2+v_2)/(2)\\79=((u_1+v_2)+(v_1+u_2))/(2)\\79=((u_1+v_2))/(2)+((v_1+u_2))/(2)\\Now,\\As\ the\ mid-point\ between\ two\ points\ can\ be\ calculated\ by\\ its\ average:(u+v)/(2) \\Hence,\\u_2\ lies\ in\ the\ mid-point\ of\ u_1\ and\ v_2.\\u_2=((u_1+v_2))/(2)\\`

`Hence,\\By\ substituting\ u_2=((u_1+v_2))/(2),\\79=u_2+(v_1+u_2)/(2)\\As\ v_1=u_2[Proven],\\79=u_2+ (u_2+u_2)/(2)\\79=u_2+(2u_2)/(2)\\79=2u_2\\Hence,\\u_2=(79)/(2)\\u_2=39.5\\Now,\ as\ we\ already\ know\ that\ the\ class\ size=5,\\v_2-u_2=5\\Hence,\\v_2=5+u_2\\Here,\\v_2=5+39.5\\v_2=44.5\\Hence,\\The\ upper\ limit\ of\ the\ second\ interval=44.5\\The\ lower\ limit\ of\ the\ second\ interval=39.5\\`