A survey of adults found that 34% say their favorite sport is professional football. You randomly select 130 adults and ask them if their favorite sport is professional football. a) Find the probability that at most 66 people say their favorite sport is professional football. The steps I have taken to solve this include: n=130,p=.34,q=.66 mean=np=44.2;standard deviation=sqrt(npq) =5.4 So z=(66−44.2)/(5.4)=4.03. This z-score is too high to use on my chart. I don't know where to go from here. The next question I have is how I would find the probability that MORE than 31 people say their favorite sport is professional football. I know I am doing something wrong, can anyone help? Thank you!

mksfmasterio

mksfmasterio

Answered question

2022-09-09

A survey of adults found that 34% say their favorite sport is professional football. You randomly select 130 adults and ask them if their favorite sport is professional football.
a) Find the probability that at most 66 people say their favorite sport is professional football.
The steps I have taken to solve this include:
n = 130 , p = .34 , q = .66
mean=np=44.2;standard deviation= mean = n p = 44.2 ; standard deviation = n p q = 5.4
So
z = 66 44.2 5.4 = 4.03.
This z-score is too high to use on my chart. I don't know where to go from here.
The next question I have is how I would find the probability that MORE than 31 people say their favorite sport is professional football. I know I am doing something wrong, can anyone help? Thank you!

Answer & Explanation

Zayden Dorsey

Zayden Dorsey

Beginner2022-09-10Added 18 answers

Let's assume that each adult answers independently. Let X denote an indicator random variable, equal to 1, if an adult names the professional football his favorite sport, and 0 otherwise. We take the survey's result to mean that P(X=1)=p=0.34, and P(X=0)=1−p=0.66. Thus X is a Bernoulli random variable.
Asking the question of 130 adults, and counting the total of those favoring professional football, gives
N = X 1 + X 2 + + X 130
The sum of independent Bernoulli random variables follows a binomial distribution, with parameters n=130 and p=0.34.
Thus
P ( N 66 ) = m = 0 66 ( 130 m ) ( 1 p ) 130 m p m = 0.99971
and
P ( N > 31 ) = m = 32 130 ( 130 m ) ( 1 p ) 130 m p m = 0.99168
Using the central limit theorem approximation to the above probabilities gives a good agreement. Indeed, μ = 130 p = 44.2 and σ = 130 p ( 1 p ) = 5.4, so
P ( N 66 ) Φ ( 66 44.2 5.4 ) = Φ ( 4.036 ) = 0.99973
and
P ( N > 31 ) 1 Φ ( 31 44.2 5.4 ) = 1 Φ ( 2.444 ) = 0.992736

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