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Answer:
No. of blue chips contain in a bag = P = 27.5 % = 0.27
No. of chips random selected at a time with replacement = n = 5
We know that the mean and standard deviation of the sampling distribution of the sample proportion (p) is given by :-
\mu\-_pμ\-p = p
= 0.275
\sigma_pσp = \sqrt{\frac{p (1-p)}{n} }np(1−p)
Where p= Population proportion and n = sample size.
Let p be the proportion of blue chips.
As per given , we have
p= 0.275
n= 5
= \sqrt{\frac{0.275 ( 1 - 0.275)}{5} }50.275(1−0.275)
= 0.1997