Question

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.

Answer 1

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.

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) = (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:

`√(V(X)) = √(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 = (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 = √(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.