Find the value of p, if the mean of the following distribution is 18 :

Explanation

To compare man, we prepare the following table

Thus, the required mean
` barx=(sumf_(i)x_(i))/(sumf_(i))`
`rArr18=(399+5p^(2)+100)/(23+5p)`
`rArr 399+5p^(2)+100pp=414+90p`
`rArr 5p^(2)+10p-15=0`
`rArr p^(2)+2p-3=0`
`rArr (p+3)(p-1)=0`
`rArrp=1 or -3`
But frequency cannot be negative, so we negative, so we reject p=-3
Hence, p=1