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.
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
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.
"\\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
"\\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"
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