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 =
b) what is the standard deviation of x? σx = (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)?

Explanation:

Part A: Mean of \( X \)

Step 1: Identify distribution and parameters

\( X \) follows a binomial distribution \( X \sim \text{Binomial}(n = 1100, p = 0.26) \). The mean of a binomial distribution is \( \mu = np \).

Step 2: Calculate the mean

\( \mu_X = n \times p = 1100 \times 0.26 \)
\( \mu_X = 286 \)

Step 1: Recall binomial standard deviation formula

The standard deviation of a binomial distribution is \( \sigma = \sqrt{np(1 - p)} \).

Step 2: Substitute values

\( n = 1100 \), \( p = 0.26 \), so \( 1 - p = 0.74 \).
\( \sigma_X = \sqrt{1100 \times 0.26 \times 0.74} \)
First, calculate \( 1100 \times 0.26 \times 0.74 = 1100 \times 0.1924 = 211.64 \)
Then, \( \sigma_X = \sqrt{211.64} \approx 14.55 \) (rounded to 2 decimal places)

Step 1: Check normal approximation conditions

\( np = 1100 \times 0.26 = 286 \geq 5 \) and \( n(1 - p) = 1100 \times 0.74 = 814 \geq 5 \), so normal approximation is appropriate.

Step 2: Apply continuity correction

For \( P(268 \leq X \leq 300) \), with continuity correction, it becomes \( P(267.5 < X < 300.5) \).

Step 3: Standardize the bounds

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 \( x = 300.5 \):
\( z_2 = \frac{300.5 - 286}{14.55} = \frac{14.5}{14.55} \approx 0.996 \approx 1.00 \) (rounded)

Step 4: Find probabilities from z-table

\( P(Z < -1.27) \approx 0.1020 \) (from z-table)
\( P(Z < 1.00) = 0.8413 \) (from z-table)

Step 5: Calculate the probability

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

Answer:

\( \mu_X = \boxed{286} \)

Part B: Standard Deviation of \( X \)