A school will offer pizza on Friday's if at least 30% of the students will buy it. A sample of 50 students asked if they would buy pizza on friday and 10 respond that they would. determine the probability of getting a sample of this size with the proportion or lower given a population with a proportion of. 30

0.0614

Step-by-step explanation:

Solution:-

- The school offers pizza on friday when at-least 30% students are willing to buy.

- The minimum required proportion of students, p = 30%

- A sample of n = 50 students were taken and out of the sample x = 10 people responded willing to buy pizza.

- We will denote a random variable X: the number of students willing to buy Pizza on fridays.

- For the sample of n = 50, for Pizza to be offered on friday the expected number of student who should opt for buying must be at-least:

Expected (Mean - u) = n*p = 50*0.3 = 15 people.

- The standard deviation for the normal distribution can be determined as follows:

standard deviation (s) = √n*p * (1-p)

= √50*0.3 * (0.7)

= √10.5 = 3.2404

- We will assume that the random variable X follows normal distribution:

X ~ Norm (15, 3.2404^2)

- To determine the probability of sample of size n = 50 with proportion or lower. The number of people willing to buy pizza on friday as per survey are x = 10.

- Evaluate the standard normal (Z-score):

P (X < x) = P (Z < (x - u) / s)

P (X < 10) = P (Z < (10 - 15) / 3.2404)

P (X < 10) = P (Z < - 1.54301)

- Use the standard normal tables to determine the value of P (Z < - 1.54301):

P (X < 10) = P (Z < - 1.54301) = 0.0614

Answer: The probability of getting a sample with the proportion or lower given a population with a proportion of. 30 is 0.0614.