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.

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:

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, = 25, = 3.536

Using the z test statistics;

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

b, i believe, but not sure.

1 1/7 hope i

step-by-step explanation:

8/7 = (1 * 7 + 1) / 7 = 1 + 1/7 = 1 1/7