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=0.95, n=255"
The mean value and standard deviation of a binomial random variable "X" are
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=255(0.95)=242.25"
"\\sigma=\\sqrt{np(1-p)}=\\sqrt{255(0.95)(1-0.95)}\\approx3.4803"
a. Use Normal approximation to determine the binomial probability of at least 246 passengers showing up.
"=1-P(Z\\leq{245.5-242.25 \\over 3.4803})\\approx1-P(Z\\leq0.933828)\\approx"
"\\approx1-P(Z\\leq0.933828)\\approx1-0.82480357\\approx"
The binomial probability of at least 246 passengers showing up is "0.1752"
(b) Since "17.52 \\%>5\\%," then the airline change (decrease) the number of tickets it sells for the flight.
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