Answer)
Given n = sample size = 100
P = 0.16 (16%)
Now first we need to check the conditions or normality that is if n*p and n*(1-p) both are greater than 5 or not
N*p = 16
And N*(1-p) = 84
As both are greater than 5, conditions are met and we can use standard normal z table to estimate the answers
A)
Standard error is = √{p*(1-p)}/√n
= 0.03666060555
B)
We need to find
P(10<x<25)
P(0.1<x<0.25)
= p(x<0.25) - p(x<0.1)
P(x<0.25)
Z = (x-mean)/s.E = (0.25-0.16)/0.03666060555
Z = 2.45
From z table, p(z<2.45) = 0.9929
P(x<0.1)
Z = (0.1-0.16)/0.03666060555
Z = -1.64
From z table, P(Z<-1.64) = 0.0505
Required probability is 0.9929-0.0505 = 0.9424
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