Question

50 students live in a dormitory. The parking lot has the capacity for 30 cars. Each student has a car with probability 12 , independently from other students. Use the CLT (with continuity correction) to find the probability that there won't be enough parking spaces for all the cars.

Answer 1

Answer:

P(X is greater than 30) = 0.06

Step-by-step explanation:

Given that:

Sample proportion (p) = 0.5

Sample size = 30

The Binomial can be approximated to normal with:

$\mu = np = 50 \times 0.5 \\ \\ \mu= 25$

$\sigma = \sqrt{np(1-p) } \\ \\ \sigma = \sqrt{50 \times (0.5)(1-0.5) } \\ \\ \sigma = 3.536$

To find:

P(X> 30)

So far we are approximating a discrete Binomial distribution using the continuous normal distribution. 30 lies between 29.5 and 30.5

Normal distribution:

x = 30.5, $\mu$ = 25, $\sigma$ = 3.536

Using the z test statistics;

$z = \dfrac{x - \mu}{\sigma}$

$z = \dfrac{30.5 - 25}{3.536}$

$z = \dfrac{5.5}{3.536}$

z = 1.555

The p-value for P(X>30) = P(Z > 1.555)

The p-value for P(X>30) = 1 - P (Z< 1.555)

From the z tables;

P(X> 30) = 1 - 0.9400

Thus;

P(X is greater than 30) = 0.06