We need to find out how many adults must the brand manager survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage.
From the given data we know that our confidence level is 90%. From Standard Normal Table we know that the critical level at 90% confidence level is 1.645. In other words, 
.
We also know that E=5% or E=0.05
Also, since, 
is not given, we will assume that =0.5. This is because, the formula that we use will have 
in the expression and that will be maximum only when =0.5. (For any other value of , we will get a value less than 0.25. For example if, is 0.4, then 
and thus, 
.).
We will now use the formula
We will now substitute all the data that we have and we will get
which can approximated to n=271.
So, the brand manager needs a sample size of 271
<h2>A)</h2>
<h2>B)</h2><h3>Since we only have 75 tickets to sell, the domain is:</h3><h2>t ε [ 0 , 75 ]</h2>
<h2>Range: (extra)</h2><h2>m ε [ 0 , 225 ] </h2>
La fille a mudane guah garcon la homme la femme. 6/9
Answer:
34.87% probability that all 5 have a wireless device
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
81% of students own a wireless device.
This means that 
If 5 students are selected at random, what is the probability that all 5 have a wireless device?
This is P(X = 5) when n = 5. So
34.87% probability that all 5 have a wireless device
