I am having some difficulty with the following question from my textbook. I have really been trying to understand the use of normal and binomial approximations, but I'm getting really confused. Any help is appreciated!

During an advertising campaign, the manufacturers of Wolfitt (a dog food) claimed that 60% of dog owners preferred to buy Wolfitt.

Assuming that the manufactures claim is correct for the population of dog owners, how would I a) calculate using a binomial distribution and b) calculate using a normal approximation to the binomial, the probability that at least 6 of a random sample of 8 dog food owners prefer to buy Wolfitt.


ad a)

The random variable is binomial distributed: $X\sim Bin(8,0.6)$

Thus the calculation is: $P(X \geq 6)=\sum_{k=6}^8 {8 \choose k} \cdot 0.6^k\cdot 0.4^{8-k}$

ad b) The formula for the approximation is:

$P(X \leq x)\approx\Phi\left( \frac{x+0.5-\mu}{\sigma} \right)$

$\Phi(z)$ is the standard normal function. The value for $\Phi(z)$ can be looked up in a table of a standard normal distribution.

$\mu=n\cdot p, \ \sigma=\sqrt{n\cdot p\cdot (1-p)}$

X is a discrete random variable. Thus $P(X \geq x)=1-P(X\leq x-1)$

$P(X \geq 6)=1-P(X \leq 5)\approx 1-\Phi\left( \frac{5+0.5-\mu}{\sigma} \right)$

I think you can go on from here.

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