Question

in a doctors office the probability of a "no show" patient for any checkup appointment on any given day is 1 out of 10. suppose that there are 18 appointments scheduled for one day. find the probability that fewer than 4 patients don't show. Pleas include how to solve on ti-84

Answer 1

Here, we are given the prior probability as: P( no show ) = 1/10 = 0.1

Out of 18 appointments scheduled for one day, the number of patients who dont show up could be modelled here as a binomial distribution given as:

Now the required probability here is computed in Ti-84 calculator as:

-- We click on the 2nd button on the calculator and select the distribution: Binomcdf in this case.

-- Then we enter the number of trials as 18 and the value of p as 0.1

-- Also we enter the x value here as 3 and then ENTER

-- We get the result as 0.9018

Therefore 0.9018 is the required probability here.