Question

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

Answer 1

Answer:

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.