Question

1. A survey of 3300 people asked them whether they liked blueberry pie. Suppose that the people were all chosen independently with a 99.98% chance of liking blueberry pie. (a) Why is the normal approximation not appropriate in this instance? (b) Calculate the probability that all 3300 people will say they like blueberry pie. (c) Use a Poisson approximation to find the probability that 3297 of the people will say that they like blueberry pie.

Answer 1

a

normal approximation not appropriate in this instance because

`np = 3300 * 0.9998 = 3299.3 > 5`

and

`nq = 3300 * 0.0002 = 0.66 < 5`

b

The probability is
`P(X = 3300) = 0.52`

c

The probability is
`P(X = 3287) = 0`

Explanation:

From the question we are told that

The sample size is
`n = 3300`

The chance of liking blueberry pie is
`p = 0.9998`

The chance of not liking blueberry pie is
`q = 0.0002`

For normal approximation is possible if

`np > 5` ,
`nq > 5`

Now let test

`np = 3300 * 0.9998 = 3299.3 > 5`

and

`nq = 3300 * 0.0002 = 0.66 < 5`

The probability that all 3300 people will say they like blueberry pie is mathematically represented as

`P(X = 3300) = p^(3300)`

`P(X = 3300) = (0.9998)^(3300)`

`P(X = 3300) = 0.52`

The probability that 3297 of the people will say that they like blueberry pie is mathematically represented as

`P(X = 3287) = ((np)^(3297! ) * e^(-np))/(3297)`

`P(X = 3287) = (0)/(\infty)`

`P(X = 3287) = 0`