If the mean of the distribution is 2.6, then the value of y is
`{:("Variate x",1,2,3,4,5),("Frequency f of x",4,5,y,1,2):}`

If the mean of the distribution is 2.6, then the value of y is {:("Variable x",1,2,3,4,5),("Frequency f of x",4,5,y,1,2):}

If the mean of the following distribution is 2.6, then the value of y is Variable (x) : 1 2 3 4 5 Frequency (y) : 4 5 y 1 2 (a) 3 (b) 8 (c) 13 (d) 24

If the mean of the following distribution is 2.6, then the value of y is Variable (x) : 1 2 3 4 5 Frequency (y) : 4 5 y 1 2 (a) 3 (b) 8 (c) 13 (d) 24

If the mean of the following distribution is 2.6 Then the value of p is Variable x: 1 2 3 4 5 frequency of x: 4 5 P 1 2

If the mean of the following distribution is 2.6., then y = . . . . {:(x_(i),1,2,3,4,5),(f_(t)"",4,5,y,1,2):}

If the mean of the following distribution is 6, find the value of p {:(x:, 2, 4, 6, 10, p+5), (f:, 3, 2, 3, 1, 2):}

If the mean of the following distribution is 6, find the value of p. x: 2 4 6 10 p+5 f: 3 2 3 1 2

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 1.46. {:(No. of accident (x), 0, 1 , 2 , 3, 4, 5 , Total), (Frequency (f): , 46 , ? , ? , 25 , 10 , 5 , 200):}

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 1.46. No. of accident (x), 0, 1 , 2 , 3, 4, 5 , Total Frequency is 200 and no. of days (f): 46 , ? , ? , 25 , 10 , 5