Question

Let r be the binomial random variable corresponding to the number of people that will live beyond their 90th birthday,

r ≥ 15.


We want to find
P(r ≥ 15)
using the normal approximation given 625 trials and a probability of a 4.4% success on a single trial.

Answer 1

P(r ≥ 15) = 0.9943.

Explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

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 z-score 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 p-value, 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))`.

625 trials and a probability of a 4.4% success on a single trial.

This means that
`n = 625, p = 0.044`

Mean and standard deviation:

`mu = E(X) = np = 625*0.044 = 27.5`

`\sigma = √(V(X)) = √(np(1-p)) = √(625*0.044*0.956) = 5.13`

P(r ≥ 15)

Using continuity correction, this is
`P(r \geq 15 - 0.5) = P(r \geq 14.5)`, which is 1 subtracted by the p-value of Z when X = 14.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (14.5 - 27.5)/(5.13)`

`Z = -2.53`

`Z = -2.53` has a p-value of 0.0057

1 - 0.0057 = 0.9943

So

P(r ≥ 15) = 0.9943.