Question

Zener cards are often used to test the "psychic ability" of individuals. In the Zener deck, there are five different patterns displayed, and each has a 1/5 probability of being drawn from a well-shuffled deck. The five patterns are: circle, plus sign, wavy lines, empty box, and star. One hundred trials were conducted and your very impressive friend guessed right on 41 of those trials. Given this sample, can we use the normal approximation to the binomial?

Answer 1

We need to check the conditions in order to use the normal approximation.

`np=100*0.2=20 \geq 10`

`n(1-p)=100*(1-0.2)=80 \geq 10`

If we check the conditions with the estimated proportion we got:

`n\hat p=100*0.41=41 \geq 10`

`n(1-\hat p)=100*(1-0.41)=59 \geq 10`

So we see that we satisfy the conditions and then we can apply the approximation.

Explanation:

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, on this case we now that:

`X \sim Binom(n=100, p=0.2)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

We need to check the conditions in order to use the normal approximation.

`np=100*0.2=20 \geq 10`

`n(1-p)=100*(1-0.2)=80 \geq 10`

If we check the conditions with the estimated proportion we got:

`n\hat p=100*0.41=41 \geq 10`

`n(1-\hat p)=100*(1-0.41)=59 \geq 10`

So we see that we satisfy the conditions and then we can apply the approximation.