Answer:
0.0614
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.