Question

if the mode of the following distribution is 17 2/3 find p.

0-5 5-10 10-15 15-20 20-25 25-30 30-35
6 11 p 24 17 13 5

Answer 1

Sorry We aren't able to understand your question.

Answer 2

Answer:

31

Given:-

$distribution \: \: \: \: \: \: \: \: \: \: \: \: \: \: frequency \\ 0 - 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 6 \\ 5 - 10 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 11 \\ 10 - 15 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p \\ 15 - 20 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 24 \\ 20 - 25 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 17 \\ 25 - 30 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 13 \\ 30 - 35 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 5$

To find:-

• missing frequency P

Solution:-

• the mode of the following distribution is 17 2/3 find p.

now,

Here..

maximum frequency = 24,

hence, modal class is 15 - 20

$l_{1} = 15 \\ f_{1} = 24 \\ f_{2} = 17\\ f_{0} = p \\ h = 5 \\ \\ a s\: we \: all \: \: know \\ \\ mode = l_{1} + \frac{f_{1} - \: f_{0} }{2 \:f_{1} - f_{0} - \: f_{2}} \times h$

here

mode = 17 2/3

$mode = l_{1} + \frac{f_{1} - \: f_{0} }{2 \:f_{1} - f_{0} - \: f_{2}} \times h \\ \\ \: 17 \frac{2}{3} = 15 + \frac{24 - p}{2 \times 24 - p - 17} \times 5 \\ \\ \frac{53}{3} = 15 + \frac{24 - p}{48 - p - 17} \times 5 \\ \\ \frac{53}{3} = 15 + \frac{24 - p}{31 - p} \times 5 \\ \\ \frac{53}{3 \times 5} = \frac{465 - 15p + 24 - p}{31 - p} \\ \\ \frac{53}{15} = \frac{465 - 16p + 24}{31 - p} \\ \\ \frac{53}{15} = \frac{489 - 16p}{31 - p} \\ \\ 53(31 - p) \: = 15(498 - 16p) \\ \\ 1643 - 53p = 7470 - 240p \\ \\ - 53p + 240p = 7470 - 1643 \\ \\ 187p = 5827 \\ \\ p = \frac{5827}{187} \\ \\ p = 31 \: \: aprox$