8.

Mathematics

1 answer:

borishaifa [10]3 years ago

6 0

Answer: 143

Step-by-step explanation: So, first, you add the two first amounts, which are $12 and $22, which is 34. Then, you divide the amount you have (4882) by the amount you need (34). And finally, you take away the decimal, and you have your answer!

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For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

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