Final answer:
The probability of getting less than 63 sales for the mail order company can be approximated using the normal approximation to the binomial distribution with mean 66.77 and standard deviation 7.57. A Z-score is calculated using a continuity correction factor, and the probability is found using a standard normal table or a calculator.
Step-by-step explanation:
To find the probability of the mail order company getting less than 63 sales when mailing advertisements to 607 people with an 11% success rate, we can model this scenario with a binomial distribution, where n=607 and p=0.11. For a discrete random variable X representing the number of sales, the probability of getting less than 63 sales is P(X<63). However, due to the large sample size, we can use a normal approximation to the binomial distribution. The mean (μ) of the distribution is n*p, and the standard deviation (σ) is √(n*p*(1-p)).
First, calculate the mean (μ) and the standard deviation (σ):
μ = n*p = 607*0.11 = 66.77
σ = √(n*p*(1-p)) = √(607*0.11*0.89) ≈ 7.57
Next, convert the discrete X to a continuous Z using the continuity correction factor because we are using the normal approximation:
Z = (X - 0.5 - μ) / σ
So, for X = 62.5 (subtracting 0.5 for continuity correction), we get:
Z = (62.5 - 66.77) / 7.57 ≈ -0.56
Finally, we look up Z = -0.56 on the standard normal distribution table or use a calculator to find P(Z < -0.56). Using a TI-83, 83+, 84, or 84+ Calculator, we would find this probability which would give us P(X < 63).