In a hospital 60% of patients are dying of a disease. If on a certain day, 8 pat

Question

In a hospital 60% of patients are dying of a disease. If on a certain day, 8 patients got admitted in the hospital for that disease what are the chances at least half of them survive? Lets compute this one by one and using the TI-84 First Find the probability that 4 survive: 23.22 Find the probability that 5 survive: Find the probability that 6 survive: Find the probability that 7 survive: Find the probability that 8 survive: Combine these to find the chances at least half of them survive: Using your binomCDF (8, .6, ) = which is the opposite probability: the probability that less than half survive.

Explanation / Answer

x-be the patient dying of disease

here x follows binomial distribution with n=8 and p=0.6

probability that 5 survive=8C5 (0.6)5(0.4)3 *100=27.869184

Probability that 6 survive=8C6(0.6)6(0.4)2*100=20.901888

Probability that 7 survive=8C7(0.6)7(0.4)1*100=8.957952

Probability that 8 survive=8C8(0.6)8*100=1.679616

combine these to find the chances at least half of them survive=23.22+27.869184+20.901888+8.957952+1.679616=82.62864

Using your binomCDF(8,6,82.62864)=0.999999374

which is the opposite probability : the probability that less than half survive

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