Respuesta :
Answer:
1) Random sample. (Assumed)
2) [tex]np=100*0.05=5 < 10[/tex]
[tex]n(1-p)=100*(1-0.05)=95 \geq 10[/tex]
We don't satisfy the condition that np>10 so then we can't use the normal approximation for this case and the answer would be:
c. No, because n(p^) < 10 or n(q^) < 10.
Step-by-step explanation:
For this case we assume that the question is: "For the following question, is it appropriate to use a normal distribution to approximate a confidence interval for the population proportion?"
An insurance company wants to know the proportion of clients who have had claims within the last year. They randomly select 100 clients from a database of all clients and checks their basic information. They find that 5% of the clients had claims within the last year.
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest "number of claims the last year", on this case we now that:
[tex]X \sim Binom(n=100, p=0.05)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We need to check the conditions in order to use the normal approximation.
1) Random sample. (Assumed)
2) [tex]np=100*0.05=5 < 10[/tex]
[tex]n(1-p)=100*(1-0.05)=95 \geq 10[/tex]
We don't satisfy the condition that np>10 so then we can't use the normal approximation for this case and the answer would be:
c. No, because n(p^) < 10 or n(q^) < 10.
