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4 years ago

8

Which of the following methods can be used to construct a pattern of 10 numbers for which it would be mathematically impossible

to predict the 11th term?
Check all that apply.

A. Make a pattern from winning lottery tickets.
B. Make a pattern, starting with 2, where each number is the next digit of e(2.71828...)
C. Make a pattern where each number is the square of the number before it, plus 7 times the cube of the number before that.
D. Make a pattern by counting the number of letters in each word of Lincoln's Gettysburg Address, starting with the first word.
E. Make a pattern by asking each of 10 friends what their favorite number is.

2 answers:

valentina_108 [34]4 years ago

8 0

Theres one more answer and its E and C

Oxana [17]4 years ago

3 0

E i think is the correct answer :-)

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b) $P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

c) $P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

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Step-by-step explanation:

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The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=1-0.2=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case we want this probability:

$P(X=14)$

And using the mass function we have this:

$P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

Part b

For this case we want this probability:

$P(X \geq 10)$

And we can find this using the complement rule:

$P(X\geq 10) = 1-P(X$

$P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}$

$P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}$

$P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}$

$P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}$

$P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}$

$P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}$

$P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}$

$P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}$

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$P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046$

And if we replace we got:

$P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

Part c

For this case we want this probability:

$P(X \leq 16)$

And we can use the complement rule like this:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115$

And if we replace we got:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

Part d

The expected value is given by:

$E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

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Step-by-step explanation:

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