Question

about 26% of the canadian population over 15 are first generation, that is, they were born outside canada. a random sample of 1100 canadians over 15 is chosen. let x = the number of first generation people in the sample.
a) what is the mean of x? μx = 286
b) what is the standard deviation of x? σx = 14.55 (round to 2 decimal places)
c) what is the probability that the sample contains between 268 and 300 first generation canadians (use the normal approximation to binomial)?
enter an integer or decimal number, accurate to at least 2 decimal places

Explanation:

Step1: Identify the distribution

We have a binomial distribution \( X \sim \text{Binomial}(n = 1100, p = 0.26) \). We use the normal approximation to the binomial, so \( X \approx N(\mu = np, \sigma^2 = np(1 - p)) \). We already know \( \mu_X = 286 \) and \( \sigma_X \approx 14.55 \) (from parts A and B).

Step2: Apply continuity correction

For \( P(268 \leq X \leq 300) \) with normal approximation, we use continuity correction: \( P(267.5 < X < 300.5) \).

Step3: Standardize the values

Calculate the z - scores for \( x = 267.5 \) and \( x = 300.5 \). The z - score formula is \( z=\frac{x-\mu}{\sigma} \).

For \( x = 267.5 \):
\( z_1=\frac{267.5 - 286}{14.55}=\frac{-18.5}{14.55}\approx - 1.27 \)

For \( x = 300.5 \):
\( z_2=\frac{300.5 - 286}{14.55}=\frac{14.5}{14.55}\approx0.996\approx1.00 \)

Step4: Find the probabilities

We need to find \( P(-1.27 < Z < 1.00) \), which is \( P(Z < 1.00)-P(Z < - 1.27) \).

From the standard normal table:
\( P(Z < 1.00)=0.8413 \)
\( P(Z < - 1.27)=0.1020 \)

Step5: Calculate the probability

\( P(-1.27 < Z < 1.00)=0.8413 - 0.1020 = 0.7393 \)

Answer:

\( 0.74 \) (rounded to two decimal places)