Mathematics, 16.07.2021 01:50 pookie879

Consider the probability that more than 97 out of 120 people will get the flu this winter. Assume the probability that a given person will get the flu this winter is 97%. Specify whether the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

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Since n(1-p) = 3.6 < 5 , 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given person will get the flu this winter is 97%.

This means that p = 0.97

120 people

This means that n = 120

Verifying the necessary conditions.

np = 120*0.97 = 116.4

n(1-p) = 120*0.03 = 3.6

Since n(1-p) = 3.6 < 5 , 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:

P(X 97) = P(X = 98) + ... + P(X = 120)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 98) = C_{120,98}.(0.97)^{98}.(0.03)^{22} \approx 0$

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.

Answer from: Quest

78.59 is the mean

step-by-step explanation:

Answer from: Quest

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