Question

in a certain hotel, 56 percent eat the free breakfast in the lobby. what is the probability that in a group of 150 ramdonly selected hotel guest more than 100 of the will eat the free breakfast in the lobby?

Answer 1

we have probability = 56/100 = 0.56

total sample size n = 150

and we have to find the probability of more than 100

Using the binomial formula for 100 or less is given as

binomcdf(n,p,largest x value), here largest x value is 100

And probability of more than 100 = 1-P(100 or less)

so, we can write it as

P(more than 100) = 1 - binomcdf(150,0.56,100) = 1 - 0.997 = 0.0030

So, the required probability that in a group of 150 randomly selected hotel guest more than 100 will eat the free breakfast in lobby is 0.0030 (rounded to four decimal places)