Answer:
probability = 0.3557
Explanation:
given data
young adults ages = 20 to 39
skip eating breakfast p = 0.238
random sample of size n = 500
to find out
we find the probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122
solution
we use here Normal Approximation to Binomial Distribution
so first consider random variable = x
so
x~ Bin (n,p) .............1
and here Normal Approximation will be
x~ Normal Approx (np, npq) .................2
so it will be
x~ (500, 0.238)
as here we know q will be
q = 1 - p
q = 1 - 0.238
q = 0.762 .............3
so
here x~ Normal Approx (119, 90.678)
and now we get P(X > 122)
so
We will convert it to Z by as that
z = ................4
and here
mean = np
and standard deviation
so here for P(X > 122)
............5
and it is P(Z>0.37)
so
probability = 1 - P(Z<0.37)
now we use here z table for value
probability = 1-0.6443
probability = 0.3557
