Respuesta :
Answer:
The approximate probability that the organization's results were within 22 percentage of the department of Tourism's results is 1
Step-by-step explanation:
Let X be the amount of tourist of the sample from Asia that philipines received. X is a binomial random varable with parameters n = 600, p = 0.63. The mean of X is np = 600*0.63 = 378 and the standard deviation is √(np(1-p)) = √(378*0.37) = 11.826. Since we are working with large values of n, we may assume that X is Normal (the results will be approximate). We want to know the probability of X being between 600*(0.63-0.22) = 246 and 600*(0.63+0.22) = 510. In order to make computations easier, we will work with the standarization of X, denoted by W, and with its cummulative density function [tex] \phi [/tex] . W is given as follows
[tex] W = \frac{X-\mu}{\sigma} = \frac{X-378}{11.826} [/tex]
Also, the values of [tex] \phi [/tex] are well know and they can be founded in the attached file. The file only has results for positive values of [tex] \phi [/tex] , however for a negative value, we can use the symmetry of the standard normal density function throught x = 0 to conclude that [tex] \phi(-x) = 1-\phi(x) [/tex] . Now we are ready to obtain the probability that we wanted
[tex]P(246<X<510 ) = P(\frac{246-378}{11.826} < X < \frac{510-378}{11.826}) = P(-11.16 < W < 11.16) =\\\phi(11.16) - \phi(-11,16) = 2 \phi(11.16)-1[/tex]
The value [tex] \phi(11.16) [/tex] is practically 1, thus 2 [tex] \phi(11.16) -1 [/tex] is also practically 1 (Maybe it wasnt 22% but a smaller gap, for big samples it shouldnt go too far from the real value almost never, check it out).
We conclude that the approximate probability that the organization's results were within 22 percentage of the department of Tourism's results is 1.
