Question

please solve question no 15​

Answer 1

Solution :

We have provided with the following distribution have a mean  of 20.2

Also , WE have to find one missing frequency it means Sigma $F_i$ is given to us .

Let us first quickly go with the  table provided to us and try to fill up all necessary details required for finding the value of P

$x_i = 10 | 15 |20|25|30$

$F_i ( Frequency) = 6|8|P|10|6$

$x_i f_i = |10 * 6 = 60|15*8 = 120|20* P = 20P|25*10= 250|30*60= 180|$

Total Sigma Fi = 30 + P

Total Sigma XiFi = 610 + 20P

Calculating mean of the distribution from direct method as the class mark is smaller and easy to calculate through it

Mean = (ΣFixi)\(ΣFi)

⇒ 20.2 = (610+20P)\(30+P)

⇒ 20.2 ( 30 + P) = 610 + 20P

⇒ 606 + 20.2P = 610 + 20P

⇒ 20.2P - 20P = 610 - 606

⇒ 0.2P = 4

⇒ 2P/10 = 4

⇒ 2P = 40

⇒ P = 40/2

⇒ P = 20

Therefore , for the above distribution the value of one missing frequency P is equal to 20

Answer 2

Step-by-step explanation:

Total Sigma Fi = 30 + P

Total Sigma XiFi = 610 + 20P

Calculating mean of the distribution from direct method as the class mark is smaller and easy to calculate through it

Mean = (ΣFixi)\(ΣFi)

⇒ 20.2 = (610+20P)\(30+P)

⇒ 20.2 ( 30 + P) = 610 + 20P

⇒ 606 + 20.2P = 610 + 20P

⇒ 20.2P - 20P = 610 - 606

⇒ 0.2P = 4

⇒ 2P/10 = 4

⇒ 2P = 40

⇒ P = 40/2

⇒ P = 20

Therefore , for the above distribution the value of one missing frequency P is equal to 20.