Question

a recently conducted survey found that 32% of customers stated that they prefer to buy items that they have seen advertised on television. suppose two hundred (200) customers were randomly selected, use an appropriate approximation to determine the probability that at most 50 of the customers prefer to buy items that they have seen advertised on television.

Answer 1

Let $X=$ the number of customers who prefer to buy items that they have seen advertised on television: $X\sim B(n, p)$

Given that $p=0.32, N=200$

The mean value and standard deviation of a binomial random variable $X$ are $\mu=np, \sigma=\sqrt{np(1-p)},$ respectively.

$$Np=200(0.32)=64>10$$ $$N(1-p)=200(1-0.32)=136>10$$

We can use Normal approximation to the Binomial

$X$ has approximately a normal distribution with $\mu=Np$ and $\sigma=\sqrt{Np(1-p)}:X\sim N(\mu, \sigma^2)$

Then

$$Z={X-\mu \over \sigma}\sim N(0, 1)$$

While the normal distribution is continuous (it includes all real numbers), the binomial distribution can only take integers. The small correction is an allowance for the fact that we’re using a continuous distribution.

Continuity Correction Factor Table

$$\begin{array}{c:c} \text{Discrete} & \text{Continuous} \\ \hline X=n & n-0.5<X<n+0.5 \\ \hline X>n & X>n+0.5 \\\hline X\leq n & X<n+0.5 \\\hline X<n & X<n-0.5 \\ \hdashline X\geq n & X>n-0.5 \end{array}$$

$$\mu=Np=200(0.32)=64$$

$$\sigma=\sqrt{200(0.32)(1-0.32)}=1.6\sqrt{17}\approx6.596969$$

Determine the probability that at most 50 of the customers prefer to buy items that they have seen advertised on television

$$P(X<50+0.5)\approx P(Z<{50.5-64 \over 6.596969})\approx$$

$$\approx P(Z<-2.046394)\approx0.0204$$

The probability that at most 50 of the customers prefer to buy items that they have seen advertised on television is approximately $0.0204$