The proportion of Canadians who eat breakfast regularly is 0.9. A simple random sample of 144 Canadians is taken.

a. What is the probability that between 0.85 and 0.875 of the sample eat breakfast regularly?
b. What is the probability that at least 0.945 of the sample eat breakfast regularly?
c. What is the probability that more than 0.945 of the sample eat breakfast regularly?
d. What is the probability between 0.065 and 0.165 of the sample DO NOT eat breakfast regularly?

To answer these questions, we will use the normal approximation to the binomial distribution since the sample size is large (n = 144) and the proportion of Canadians who eat breakfast regularly (p = 0.9) is not too close to 0 or 1.

a. To find the probability that between 0.85 and 0.875 of the sample eat breakfast regularly, we need to calculate the z-scores for both proportions and then find the area between these z-scores.

First, calculate the mean (μ) and standard deviation (σ) of the binomial distribution using the formula:
μ = n * p = 144 * 0.9 = 129.6
σ = √(n * p * (1 - p)) = √(144 * 0.9 * 0.1) ≈ 4.2

Now, calculate the z-score for 0.85:
z1 = (0.85 - μ) / σ = (0.85 - 129.6) / 4.2 ≈ -30.57

Next, calculate the z-score for 0.875:
z2 = (0.875 - μ) / σ = (0.875 - 129.6) / 4.2 ≈ -28.57

To find the probability between these z-scores, we can use a standard normal distribution table or a calculator. Using the table, we find:

P(0.85 < p < 0.875) ≈ P(-30.57 < Z < -28.57) ≈ 0

Therefore, the probability between 0.85 and 0.875 of the sample eating breakfast regularly is approximately 0.

b. To find the probability that at least 0.945 of the sample eat breakfast regularly, we need to calculate the z-score for 0.945 and find the area to the right of this z-score.

Again, calculate the z-score:
z = (0.945 - μ) / σ = (0.945 - 129.6) / 4.2 ≈ -18.92

Using the table or a calculator, we find:

P(p ≥ 0.945) ≈ P(Z ≥ -18.92) ≈ 1

Therefore, the probability that at least 0.945 of the sample eat breakfast regularly is approximately 1.

c. To find the probability that more than 0.945 of the sample eat breakfast regularly, we subtract the probability of all values less than or equal to 0.945 from 1.

P(p > 0.945) = 1 - P(p ≤ 0.945)

Using the table or a calculator, we find:

P(p > 0.945) ≈ 1 - P(Z ≤ -18.92) ≈ 1 - 0 ≈ 1

Therefore, the probability that more than 0.945 of the sample eat breakfast regularly is approximately 1.

d. To find the probability that between 0.065 and 0.165 of the sample do not eat breakfast regularly, we need to calculate the z-scores for both proportions (since we're interested in the proportion of non-breakfast eaters) and then find the area between these z-scores.

First, calculate the mean and standard deviation of the binomial distribution for the proportion of non-breakfast eaters:
μ = n * (1 - p) = 144 * (1 - 0.9) = 14.4
σ = √(n * p * (1 - p)) = √(144 * 0.9 * 0.1) ≈ 4.2

Now, calculate the z-scores for 0.065 and 0.165:
z1 = (0.065 - μ) / σ = (0.065 - 14.4) / 4.2 ≈ -3.38
z2 = (0.165 - μ) / σ = (0.165 - 14.4) / 4.2 ≈ -2.92

Using the table or a calculator, we find:

P(0.065 < p < 0.165) ≈ P(-3.38 < Z < -2.92) ≈ 0.0033

Therefore, the probability between 0.065 and 0.165 of the sample not eating breakfast regularly is approximately 0.0033.