Full Question
The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238. suppose that lance, a nutritionist, surveys the dietary habits of a random sample of size n=500 of young adults ages 20–39 in the united states.
Apply the central limit theorem for the binomial distribution to find the probability that the number of individuals, ?, in Lance's sample who regularly skip breakfast is greater than 122. You may find table of critical values helpful.
Express the result as a decimal precise to three places.
Answer:
The probability using the Normal Approximation that out of lance’s sample more than 122 people is 0.356
Explanation:
Given
n = Sample Size = 500
p = Probability = 0.238
Using the normal approximation to binomial distribution,
Let X = Event such that a person skips breakfast
If X ~ Binomial (n,p)
Using normal approximation
X ~ Normal (np,npq)
Where n = 500 and p = 0.238
So, X ~ Binomial (n,p) becomes
X ~ Binomial (500 , 0.238)
Using Normal Approximatiom
X ~ (119, 90.678)
The critical table is then constructed as follows;
Binomial --------- Normal
P(X = a) --------- P(a - 0.5 < X < a + 0.5)
P(X ≥ a) --------- P(X > a - 0.5)
P(X > a) --------- P(X > a + 0.5)
P(X ≤ a) --------- P(X < a + 0.5)
P(X < a) --------- P(X < a - 0.5)
Calculating the probability using the Normal Approximation that out of lance’s sample more than 122 people skips
This can be written as P(X > 122)
Looking at the critical table above.
P(X>a) ----- P(X>a + 0.5)
So,
P(X > 122) ---- P(X > 122 + 0.5)
P(X > 122) ---- P(X > 122.5)
Calculating Z score using
z = (x - u)/√σ²
X = 122.5
From X ~ (119, 90.678)
u =mean = 119
σ = standard deviation = √90.678
So,
Z = (122.5 - 119)/√90.678
z = 0.367550550865750
Z = 0.37 ---- Approximated
P(X > 122.5) = P(Z > 0.37)
P(X > 122.5) = 1 - P(Z<0.37) --- using z table
P(X > 122.5) = 1 - 0.6443
P(X > 122.5) = 0.3557
P(X > 122.5) = 0.356 -- Approximated
