Let X be the number of people that show up to the hotel.

X has the binomial distribution with 215 trials and a success probability of 0.90

In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(n!(n-1)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, …, n

for all guess to arrive and get a room you need 200 or fewer to show up.

P(X ≤ 200) = 1 – P(X > 200)

=1 – ( P[X = 201] + P[X = 202] + …. + P[X = 215])

= 0.95035

You can use the normal approximation to the binomial if np > 10 and n(1-p) > 10. Both are true here.

using the normal approximation to the binomial you have X ~ N(n*p, n*p*(1-p)) = N(193.5, 19.35)

P[X < 200] = P[X < 200.5] because of the continuity correction. Now convert to standard units P[X < 200.5] = P[Z < (200.5 - 193.5)/Sqrt(19.35)] = P[Z < 1.59] where Z is the standard normal P[Z < 1.59] = 0.9441 from the standard normal tables. looking at your answer you forgot the continuity correction.

Given: inhabitants propose ? = 80 mosquitoes/m², the inhabitants primary deviation is ? = 12 mosquitoes/m², pattern length is n = 9. on account which you’re instructed to anticipate that the variable is often allotted, the first commonplace distribution (as antagonistic to the student’s t distribution) could be utilized. by way of fact we are sorting out suggestions a pair of pattern propose, we use right here formulation: z = (X-bar – ?)/(?/?n). z = (80 two.8 – 80) / (12 / ?9) z = 2.8 / (12 / 3) z = 2.8 / 4 z = 0.7 by way of fact the question asks for the risk of the final being *extra* than 80 two.8, discover the section decrease than the first commonplace curve to the main suitable of z = 0.7. using an information superhighway calculator (see components decrease than), the risk is p = 0.242. (of direction you will discover this fee using a primary commonplace table from a statistics textbook besides.)