Question) It is known that, on average, 2 people in 5 in a certain country are...

Question) It is known that, on average, 2 people in 5 in a certain country are overweight. A random sample of 400 people is chosen. Using a suitable approximation, find the probability that fewer than 165 people in the sample are overweight.

It is based on using normal distribution as an approximation for the binomial distribution. Please explain in detail with all logical steps for better understanding so as to how to solve these types of questions.

Expert Solution

p=2/5=0.40

n=400

Mean = np = 160
std dev ,σ=√np(1-p)= √(0.4*0.6/400) = 9.7980

Binomial to normal approximations with continuity factor

P(X <   165   ) = P(Xnormal <   164.5   )

Z=(Xnormal - µ ) / σ =     (164.5-160)/9.798)=       0.4593

=P(Z<   0.4593   ) =    0.6770

-----------------------------------------------

Binomial to normal approximations without continuity factor


P(X < 165   )

Z=(X - µ ) / σ =        (165-160)/9.798)=       0.510

=P(Z< 0.510   ) =    0.6951

please revert for doubt

orchestra answered 1 year ago

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