Answer:
Probability = 0.3557
Explanation:
Data provided
Young adults age range = 20 to 39
Proportion skipping breakfast p = 0.238
Sample size n = 500
Objective
We aim to determine the probability that more than 122 of Lance's sampled individuals do not eat breakfast regularly.
Solution
Utilizing the Normal Approximation for the Binomial Distribution,
let's define our random variable as x:
x ~ Bin (n,p).............1
Using Normal Approximation gives us:
x ~ Normal Approx (np, npq).................2
This simplifies to:
x ~ (500, 0.238)
where we know q will be:
q = 1 - p
q = 1 - 0.238
q = 0.762.............3
Hence, we find:
x ~ Normal Approx (119, 90.678)
Now we want to compute P(X > 122).
First, we convert this to Z:
z =
................4
Where the mean is
= np
and standard deviation is
For P(X > 122, we evaluate:
............5
This results in P(Z > 0.37).
Thus,
Probability = 1 - P(Z < 0.37)
Next, we'll refer to the z table:
Probability = 1-0.6443
Probability = 0.3557