The results of a recent survey indicate that 58% of households in the United States own a gas grill. If you randomly select 100 households, what is the probability that exactly 65 households will own a gas grill?
P(65) = 100C65 (.58)65 (.42)35 = 0.0299216472
Let me show you how to use that nice calculator of yours.
Press 2ND DISTR -> :binompdf( -> ENTER
Enter the values for n, p, and r separated by commas
binompdf(100,.58,65) -> ENTER
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What is the probability that at least 65 households will own a gas grill?
P(at least 65) = P(65) + P(66) + P(67) + P(68) + P(69) + P(70) + P(71) +… + P(100)
Press 2ND DISTR -> :binomcdf( -> ENTER
binomcdf(100,.58,64) -> ENTER
CDF means cumulative density function and it allows for the computation for the probability of “r and fewer” successes. It gives you a cumulative total of probabilities. The reason I entered 64 instead of 65 is I wanted to find the probability of 0 successes through 64 successes. I want to subtract the probabilities from 0 through 64. This is the results that is on the screen – this number represents the total cumulative probabilities from 0 to 64.
Now in order to find the probability of at least 65, which is the complement of the event 64 and fewer, I will subtract this result from 1 and the solution is given.
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Example: The results of a recent survey indicate that 71% of people in the United States use more than one topping on their hotdogs. If you randomly select 250 people in the United States, what is the probability that exactly 178 of them will use more than one topping?
binompdf(250,.71,178) = 0.0555119581
What is the probability that at least 178 of them will use more than one topping?
1 – binomcdf(250,.71,177) = 0.5039008221
