Gluten sensitivity affects approximately 15% of people in the us.You decide to randomly sample 800 individuals and have them report their gluten intolerance. 96 people report gluten sensitivity.
Let we have to calculate the probability that 115 individual in the sample with wheat intolerance.
Probability of wheat intolerance is p = 0.15
Sample size, n = 800
As np = 800"\\times" 0.15 = 120 > 10 and n(1-p) = 800"\\times" (1-0.15) = 680 > 10,
we can use the normal approximation to the binomial distribution with continuity correction.
a. Let X be the number of individuals in the sample with wheat intolerance.
X~N(np, np(1-p) or X~N(120, 102)
P(X = 115) = P(114.5 < x<115.5)
"=P(\\dfrac{(114.5-120)}{\\sqrt{102}}< z< \\dfrac{(115.5-120)}{\\sqrt{102}})"
"=P(-0.5446< z< -0.4456)"
= 0.293
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