Respuesta :
Answer:
Since [tex]n(1-p) = 3.6 < 5[/tex], the normal distribution cannot be used as an approximation to the binomial probability to approximate the probability.
Using the binomial distribution, 100% probability that more than 97 out of 120 people will get the flu this winter.
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
Assume the probability that a given person will get the flu this winter is 97%.
This means that [tex]p = 0.97[/tex]
120 people
This means that [tex]n = 120[/tex]
Verifying the necessary conditions.
[tex]np = 120*0.97 = 116.4[/tex]
[tex]n(1-p) = 120*0.03 = 3.6[/tex]
Since [tex]n(1-p) = 3.6 < 5[/tex], the normal distribution cannot be used as an approximation to the binomial probability to approximate the probability. Thus, the binomial distribution has to be used.
Probability using the binomial distribution:
[tex]P(X > 97) = P(X = 98) + ... + P(X = 120)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 98) = C_{120,98}.(0.97)^{98}.(0.03)^{22} \approx 0[/tex]
Probability close to 0, but below the mean, which means that the probability of the number being above this is 100%.
Using the binomial distribution, 100% probability that more than 97 out of 120 people will get the flu this winter.
